an exceptional standing jump would raise a person 0.80 m off the ground. to do this, what force must a 68-kg person exert against the ground? assume the person crouches a distance of 0.20 m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground.
The force applied by the person is 2587.5N. The energy approach is by far and away the easiest. Earlier work equals potential energy (after).
Most issues involving forces and velocities can be solved using energy formulas like Ek = EP or W = Ep or W = Ek.
Always consider using an energy formula first; they frequently work and prevent the need for many steps to solve for one thing before moving on to another. It would take a while to find a solution if you approached it the way you did since you would have to tackle several problems as you went along. You must determine the initial takeoff speed using F = ma the way you did since you would have to tackle several problems as you went along. You must determine the initial takeoff speed using F = ma before utilizing V2f=V2i + 2ad then, given that Vf is zero, get Vi. You would then factor that into the same equation using that as your takeoff speed V2f = V2i + 2ad but this time it would be Vf. Vi would be 0, therefore the answer to the equation would be "a." The energy approach is by far and away the easiest.
Earlier work equals potential energy (after)
Fd = mghF = mgh/dF = (66)(9.81)(0.8)/0.2F= 2587.5N
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1. A car begins at a speed of 3m/s and accelerates at 2m/s² over a distance of 40m,
calculate the final speed of the car.
[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]
[tex]\qquad \tt \rightarrow \: v = 13 \:\: m/s[/tex]
____________________________________
[tex] \large \tt Solution \: : [/tex]
[tex] \texttt{Initial speed of the car (u) = 3 m/s} [/tex][tex] \texttt{final speed of the car (v) = ??} [/tex][tex] \texttt{acceleration of the car (a)= 2 m/s²} [/tex][tex] \texttt{distance covered in that time (s) = 40 m} [/tex]By third equation of motion :
[tex]\qquad \tt \rightarrow \: {v}^{2} = {u}^{2} + 2as[/tex]
[tex]\qquad \tt \rightarrow \: {v}^{2} = (3) {}^{2} + 2(2)(40) [/tex]
[tex]\qquad \tt \rightarrow \: {v}^{2} = 9 + 160[/tex]
[tex]\qquad \tt \rightarrow \: {v}^{2} = 169[/tex]
[tex]\qquad \tt \rightarrow \: v = 169[/tex]
[tex]\qquad \tt \rightarrow \: v = 13[/tex]
so, speed of the car after that time is : 13 m/s
Answered by : ❝ AǫᴜᴀWɪᴢ ❞
The time it takes to
wake up. Time interval or a time instant? (Please give me the right answer)
Answer:
Time interval
Explanation:
The amount of time it takes to wake up is a time interval because it has a start and stop time. For example, if someone's alarm goes off at 7AM and they fully wake up at 7:15AM, then the period from 7AM-7:15AM is a time interval.
A roller coaster car, 500 kg, starting at a point 80 m above the lowest point of the track. The car travels from 8 m·s -1 down the steep track. There is no friction.
9.1 What is the total energy of the roller coaster car at the top of the track?
9.2 What is the total energy of the system at the base of the rollercoaster
track?
9.3 What is the velocity of the car at a height of 30 m from
the base of the roller coaster track?
9.4 What is the velocity of the car at the base of the roller coaster track?
The total energy of the roller coaster car at the top of the track is 392,000 J.
The total energy of the system at the base of the rollercoaster track is 392,000 J.
The velocity of the car at a height 30 m above the base is 25.53 m/s.
The velocity of the car at the base of the roller coaster track is 40.4 m/s.
What is the total energy of the roller coaster car?
The total energy of the roller coaster car at the top of the track is calculated as follows;
P.E = mgh
where;
m is mass of the carg is acceleration due to gravityh is heightP.E = (500 kg) x (9.8 m/s²) x (80 m)
P.E = 392,000 J
Based on the law of conservation of energy, the total energy of the rollercoaster at the top of the track is the same as that at the bottom of the track.
Thus, total energy of the system at the base of the rollercoaster track = 392,000 J.
The velocity of the car at a height 30 m above the base is calculated as follows;
v² = u² + 2gh
v² = (8)² + 2(9.8)(30)
v² = 652
v = √652
v = 25.53 m/s
The velocity of the car at the base of the roller coaster track is calculated as follows;
v² = u² + 2gh
v² = (8)² + 2(9.8)(80)
v² = 1632
v = √1632
v = 40.4 m/s
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A massless rod is fixed to a wall on one end. A cable is attached to the wall and the center of the rod. The cable makes an angle of 60° to the wall. If the maximum tension of the cable is 10 n, what is the maximum amount of weight that can hang from the end of the rod?.
The most extreme sum of weight that can hang from the conclusion of the pole is 8.66 N.
The most extreme sum of weight that can hang from the conclusion of the bar is calculated as follows.
Apply the equation for pressure utilizing Newton's moment law of movement.
Newton's moment law of movement states that the constrain experienced by a question is specifically relative to the item of the mass and speeding up of the question.
T sinθ = W
Where T is maximum tension in cable
θ is angle of inclination of cable
W is weight of cable
10 sin (60) = W
8.66 N = W
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an object accelerates from rest to a velocity of 4.0 m/s in 8 seconds. what
was its acceleration
Answer:
Acceleration (a) = 0.5 mp[tex]s^{2}[/tex]
Explanation:
Given,
Object starts from rest, hence the initial velocity of the object is said to be 0
Velocity in the next 8 seconds is 4.0 m/s
u = 0 m/s
v = 4 m/s
t = 8 secs
We know,
a = (v-u)/t
=> a = (4-0)/8
=> a = 4/8 = 0.5 m/[tex]s^{2}[/tex]
a snowboarder glides down a 60-mm-long, 15∘∘ hill. she then glides horizontally for 10 mm before reaching a 26 ∘∘ upward slope. assume the snow is frictionless.
v ^2f-v^2i= 2a Replace the known values,
0-16^2=2 (-4.2)S, S= 256/8.4, S= 30.4m. The maximum distance a snowboarder could travel on a 25o slope was 30.4 meters.
The snowboarder travels a distance of H=50 meters on the hill. The hill slopes at a 15 degree slant. snowboarder's horizontal distance traveled after passing a slope, S=10m Crossing a horizontal distance, there is a 25-degree rising slope.
The snowboarder's initial speed at location A is u = 0 m/s. Let v be the representation of the final velocity. G = 10m/s2 is the acceleration caused by gravity. The snowboarder's acceleration at the incline is = a = 10 sin (15 degree)
The snowboarder moved 10 meters horizontally after reaching the bottom of the hill. Before she reaches the steep slope, as depicted in the illustration, her pace is unchanged. Now, at the top of the hill, where she could go the farthest, S
Vf = 0 m/sec would be the final speed.
To calculate the furthest she could go, write the equation of motion,
v ^2f-v^2i= 2a Replace the known values,
0-16^2=2 (-4.2)S, S= 256/8.4, S= 30.4m. The maximum distance a snowboarder could travel on a 25o slope was 30.4 meters.
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experiment 4: uniform circular motion introduction: in this lab, you will calculate the force on an object moving in a circle at approximately constant speed. to calculate the force you will use newton’s second law combined with the acceleration of an object moving in a circle at constant speed. you will then compare that calculated force with a measured value. you will measure the time t required for 30 revolutions of a hanging object; the object will be held in a circular path of known radius r by a horizontal spring attached to the axis of rotation. the time t for one revolution is then t/30 and the speed v of the moving object is easily calculated from distance (the circumference of the circle 2πr) divided by time: v
This given statement about uniform circular motion and distance is proved.
What is distance?The distance between two points is how long it is.
Okay, so the first three components of this issue include more physics and math. Um Ds equals R. D. Data was thus provided for section A. We have to track down S. Therefore, we just integrate both sides equally using R. D. data. Our data are provided. Therefore, the upon integration over our data is only R 0 Santa + 12 beta zeta squared for part B. We'll make sure S = V T. We combine these two by using the above equation as an equal. In order to allow it to reach its roots once positive, we can see that this is Justin's equation for data. The other drawback, of course, is
Standard units are 247 times zero, therefore switch to the nature radio at a negative six. Additionally, we are given the time, which is 74 minutes, or 4440 seconds. Thus, clogging everything. Injecting R 0 and data mm hmm into the formula for data. Therefore, we enter better Data. V also appears. Everything is therefore known. We may confidently conclude that the data is, hmm, 1.337 x 10-5 bright. or 2.13 x 10 to 4 revolutions, in terms of turns. Okay. We simply re-insert the figures for beta and tr zero V into the earlier formulae for omega and alpha in the last section of the problem. You can take pleasure in utilizing math, even by hand, or perhaps he
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consider a parallel plate capacitor having plates of area 1.7 cm2 that are separated by 0.022 mm of neoprene rubber. you may assume the rubber has a dielectric constant κ
The simplest type of capacitor is the parallel plate capacitor. The capacitance value in Farads is fixed by the surface area of the conductive plates and the space between them, and it can be built using two metal or metallized foil plates placed parallel to one another at a distance.
Chloroprene is polymerized to create the family of synthetic rubbers known as neoprene. Good chemical stability and flexibility are displayed by neoprene over a broad temperature range.
A parallel-plate capacitor's capacitance is calculated by multiplying the dielectric constant by the permittivity of free space, the area of the plates, and the distance between them.
The generalised formula for a parallel plate capacitor's capacitance is given as C = ε(A/d), where stands for the absolute permittivity of the material being utilised as the dielectric. The value of the dielectric constant, commonly referred to as the "permittivity of open space," is 8.854 x 10-12 Farads per metre.
Therefore, Capacitance = ε x (area of plates / thickness of rubber)
= 8.854 x 10^-12 *(1700 / 0.022)
= 68.4173 farads
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water (2990 g ) is heated until it just begins to boil. if the water absorbs 5.11×105 j of heat in the process, what was the initial temperature of the water?
According to the given statement the initial temperature of the water is 54.171°C
What is unit of temperature?The SI unit of temperature is Kelvin, which is represented by the letter K. Kelvin is part of the International System of Units.
Briefing:Mass of water
= 2290 g
= 2.29 Kg
Heat absorbed (Q) = 5.11 x 10⁵ J = 511 KJ
Water is boiling, final temperature of water
= 100°C = 373 K
When water is getting warmed, the amount of heat absorbed to increase the temperature of water by 1°C is given by heat capacity.
Specific heat capacity of water
= 4.87 KJ/Kg
Mathematically,
Q = mCΔT
where
Q = Heat absorbed
m= mass of the water
C = specific heat capacity of water
ΔT = Temperature change
= Final temperature - initial temperature
Putting numerical values
511 = 2.29 x 4.87 x ΔT
⇒ΔT = 45.829
⇒Final temperature - Initial temperature = 45.829°C
⇒100°C - Initial temperature = 45.829°C
⇒ Initial temperature = 100 - 45.829 = 54.171°C
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what is the maximum number of 60-w light bulbs you can connect in parallel in a 100-v home circuit without tripping the 30-a circuit breaker? please show solution not only answer.
Answer:
50 sixty watt bulbs max
Explanation:
Watts = amps * volts
30 a * 100 v / 60 w/bulb = 50 bulbs
If u move 50 meters in 10 sec, what is yor speed?
each with a mass of 1.5 kg, move toward one another. a). If the cart moving left is traveling at 10 m/s and the cart moving right is traveling at 6 m/s, what is the magnitude and direction of the total momentum of the system? b. What is the total momentum of the system if the two carts have the same speed?
The magnitude and direction of the total momentum of the system is 6 kg m/s to the left.
The total momentum of the system if the two carts have the same speed is 0 kgm/s
What is the total initial momentum of the system?The magnitude and direction of the total momentum of the system is calculated as follows;
Pi = m1u1 + m2u2
where;
Pi is the total initial momentum of the systemm1 is mass of first objectm2 is mass of second objectu1 is initial velocity of the first object'u2 is the initial velocity of the second objectPi = (1.5 x 6) - (1.5 x 10)
Pi = - 6 kg m/s
What is the total momentum of the system at same speed?The total momentum of the system if the two carts have the same speed is calculated as follows;
Pf = (1.5 x 6) - (1.5 x 6)
Pf = 0 kg m/s
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1. Two particles are placed in a line 90.0 cm apart. One particle, q1, has a charge of 1.0 μC and the
other, q2, 3.0 μC. How far from q1 should a third charge of 2.0 μC be placed, so that the forces
acting on it are equal? HELPP
a battery with voltage v is connected to a chain of three capacitors in series each with a capacitance of , after a long time has passed each capacitor has a charge of q
A capacitor acts as a low-pass filter to increase damping and temporarily store energy, much like a spring-loaded shock absorber, to reduce vibrations when driving on washboard roads.
A battery with voltage v is connected to a chain of three capacitors in series each with a capacitance of , after a long time has passed each capacitor has a charge of q
Since there is only one channel for the charging current (iC) to take when capacitors are connected in series, the charging current (iC) going through all capacitors is the SAME.
Then, as iT = i1 = i2 = i3 etc., all capacitors connected in series have the same current flowing through them. As a result, regardless of capacitance, each capacitor will store the same quantity of electrical charge, Q, on its plates.
As we know,
Q = C*V
Applying Kirchhoff’s Voltage Law,
V = Qt/C1 + Qt/C2 + Qt/C3
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What did you observe about each subsequent for every bounce from every height?
The ball will rise back to a height equal to the initial height multiplied by the amount of kinetic energy it had before it started to ascend again (1-fraction of energy lost).
Why does a ball from a higher height bounce higher?All of the ball’s kinetic energy must be dissipated when it strikes the ground. The bounce increases with potential energy, kinetic energy, and kinetic energy alone, thus the more energy that returns to the ball and gives it more power to spring back into the air, the higher the bounce.
Ball bouncing heightThe height from which a ball is dropped, the material of the ball (and, if it is inflated, the pressure at which it is inflated), and the material of the surface from which the ball bounces determine the height to which it will rebound.
If you are measuring the height of the ball from its bottom to the ground, the radius of the ball doesn't actually matter.
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A cliff that is known to be 20 m high has a river that is 30 m wide, that runs below the cliff. With what minimum speed should you drive over the cliff so that the car crosses the river?
The minimum speed you should drive over the cliff so that the car can cross the river is 15 m/s
How to determin the minimum speedwe'll begin by obtainig the time taken for the car to cross the river. This can be obtained as follow:
Height (h) = 20 mAcceleration due to gravity (g) = 9.8 m/s²Time (t) = ?h = ½gt²
20 = ½ × 9.8 × t²
20 = 4.9 × t²
Divide both side by 4.9
t² = 20 / 4.9
Take the square root of both side
t = √(20/ 4.9)
t = 2 s
Finally, we shall determine the minimum speed as follow:
Horizontal distance (s) = 30 mTime (t) = 2 sMinimum speed (u) = ?s = ut
30 = u × 2
Divide both sides by 2
u = 30 / 2
u = 15 m/s
Thus, we can conclude that the minimum speed needed is 15 m/s
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The minimum speed should be 15 m/s to drive over the cliff so that the car will cross the river.
What is Speed?The speed of an item, which is a scalar quantity in everyday usage and kinematics, can be defined as the size of the change in position per unit of time or the size of the change in position over time for an object.
The given values in the question are :
Height of the cliff, h = 20 m
The acceleration due to gravity, g = 9.8 m/s².
Use the formula,
h = 1/2gt²
20 = 1/2(9.8)t²
20 = 4.9 t²
t = 2 seconds.
Now, calculate the minimum speed,
s = ut
30 = u × 2
u = 30/2
u = 15 m/s.
Hence, the minimum speed required by the car to cross the river is 15 m/s.
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The following graphs show average temperature data for two locations. Both locations have the same elevation. What is most likely true about the locations?
1080AQ5
A. Location A is farther from the equator than location B.
B. Location A is closer to the equator than location B.
C. Location B is located farther inland than location A.
D. Location A is in the northern hemisphere, and location B is in the southern hemisphere.
The statement that is most likely true about the locations is that location A is farther from the equator than location B ( A )
The average temperature throughout the year is higher in location B then that of location A. The temperature is typically warmer near the equator and is typically cooler near the polar regions.
But some factors such as elevation, precipitation and ocean currents might affect the climate pattern. This change in temperature between equator and polar regions is due to the fact that equatorial regions receive more light and energy from the Sun than the polar regions.
Therefore, the statement that is most likely true about the locations is that location A is farther from the equator than location B ( A )
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a movable bracket is held at rest by a cable attached at e and by frictionless rollers. knowing that the width of post fg is slightly less than the distance between the rollers, determine the force exerted on the post by each roller when α
The force exerted on the post by the roller,
A = D = 0B = 868 N ; C = 126.1 NThe equilibrium for bracket,
∑ [tex]F_{y}[/tex] = 0
T sin 20° - 270 = 0
T = 789.43 N
[tex]T_{x}[/tex] = ( 789.43 ) cos 20°
[tex]T_{x}[/tex] = 741.82 N
[tex]T_{y}[/tex] = ( 789.43 ) sin 20°
[tex]T_{y}[/tex] = 270 N
The 270 N force and [tex]T_{y}[/tex] forms a couple
τ = ( 270 ) 0.25
τ = 67.5 N m Clockwise
∑ [tex]M_{B}[/tex] = 0
( 741.82 * 0.125 ) - 67.5 + [tex]F_{CD}[/tex] ( 0.2 ) = 0
[tex]F_{CD}[/tex] = - 126.138 N
∑ [tex]F_{y}[/tex] = 0
[tex]F_{AB}[/tex] - 126.138 - 741.82 = 0
[tex]F_{AB}[/tex] = 867.96 N
The direction of force [tex]F_{CD}[/tex] is towards left and the direction of force [tex]F_{AB}[/tex] is towards right. The force [tex]F_{CD}[/tex] is exerted by roller B and the force [tex]F_{AB}[/tex] is exerted by roller C. Rollers A and D exert no force. The force exerted by the rollers exert oppositely on the post. So the force exerted by roller C is 126.1 N and the force exerted by roller B is 867.96 N.
Therefore, the force exerted on the post by the roller,
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A vibrating guitar string makes very little sound if it is not mounted on the guitar body. Why does the sound have greater intensity if the string is attached to the guitar body? (a) The string vibrates with more energy. (b) The energy leaves the guitar at a greater rate. (c) The sound power is spread over a larger area at the listener's position. (d) The sound power is concentrated over a smaller area at the listener's position. (e) The speed of sound is higher in the material of the guitar body. (f) None of these answers is correct
The sound has greater intensity if the string is attached to the guitar body as the sound power is spread over a larger area at the listener's position.
A sound wave is basically delivered by a vibrating object. As a guitar string vibrates, it sets encompassing air particles into vibrational movement. The frequency at which these air particles vibrate is rise to the frequency of vibration of the guitar string. The back and forward vibrations of the surrounding air particles make a pressure wave that voyages outward from its source. This pressure wave comprises compressions and rarefactions. The compressions are regions of high pressure, where the air particles are compressed into a little region of space. The rarefactions are districts of low pressure, where the air particles are spread and separated. This substituting design of compressions and rarefactions is known as a sound wave. A guitar string vibrating by itself does not deliver a really loud sound. The string itself disturbs exceptionally small air since its little surface zone makes exceptionally small contact with encompassing air particles. In any case, in case the guitar string is joined to a bigger object, such as a wooden sound box, at that point more air is aggravated. The guitar string forces the sound box to start vibrating at the same frequency as the string. The sound box in turn powers surrounding air particles into vibrational movement. Since of the huge surface range of the sound box, more discuss particles are set into vibrational movement. This produces a more capability of hearing a sound.
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explain and justify what would happen to the work done by the cable and magnitude of tension force as the crate was lifted the same height h but with a longer ramp.
Friction comes in two flavors: kinetic and static. Static friction affects a system or an object at rest, whereas kinetic friction affects an object in motion.
Typically, the maximal static friction between the objects is greater than the kinetic friction. Consider attempting to move a large crate across a concrete floor. The crate might not budge no matter how much you push on it. In other words, the static friction reacts to your actions by increasing to match and move in the opposite direction of your push.
However, if you exert enough force, the crate eventually looks to slip and begin to move. The kinetic friction force is lower than the static friction force, making it simpler to maintain motion once it is underway.
You would have to push more harder to initiate and maintain motion if you added mass to the crate (by, say, stacking a box on top of it). On the other hand, if you greased the concrete, it would be simpler to start and maintain the box.
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John rode 3,150 m at an average speed of 350 m/min. if he had ridden at average of 375 m/min instead, how many seconds fewer would the ride have taken him to complete?
John will take 0.6 min or 36 seconds more faster during uniform motion than before.
We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
s = v . t
where v is velocity, t is time and s is displacement.
From the question above, we know that
v1 = 350 m/min
v2 = 375 m/min
s = 3150 m
Find the initial time
s = v1 . t
3150 = 350 . t
t = 9 min
Find second time
s = v2 . t
3150 = 375 .t
t = 8.4 min
John will take 0.6 min or 36 seconds more faster than before
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1. A scientist wants to test the sleeping effects of a mild animal tranquilizer on two of his pet rats. He decides that he will put them in two cages of the same size and shape and place them in the same area of his lab for 24 hrs. He decides that he is going to turn the lights on at 5am and turn the lights of at 8pm each day. He feeds both rats lunch each day at noon and chops up 5mg of animal tranquilizer into their food. He notes that at about 5pm both rats fall asleep and wake up at about 5am the following day. He continues this procedure for 15 days and notes the same pattern each day. He concludes that his study shows that the animal tranquilizer will cause all animals to fall asleep at 5pm for a total of 12 hrs.
Did this experiment give accurate results? Why or why not?
Since both rats get the tranquilizer and there is no control condition, the experiment is not accurate.
An experiment is what?An experiment would be carried out by a scientist in search of data and knowledge. We are aware that in an experiment, the impact of the independent variable on the dependent variable must be determined.
In this instance, the dosage of the medication is the independent variable, and the duration of sleep time for the pet rats is the dependent variable.
Let's notice that a controlled setting is required for every experiment.
What would be used to validate the experiment as we now understand it is the control setting of the experiment?We can observe that this experiment lacks a controlled environment, proving that it is not reliable.
A control setting should be introduced, and this experiment would need to be evaluated.
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Stephen learned that any two objects exert a gravitational force on each other. If the distance between two objects triples, the gravitational force between them will change by a factor of what?.
The gravitational force between them will change by a factor 1/9.
We need to know about gravitational force to solve this problem. The gravitational force is the force caused by two masses objects. The magnitude of gravitational force can be determined as
F = G.m1.m2 / R²
where F is the gravitational force, G is the gravitational constant (6.674 × 10¯¹¹ Nm²/kg²), m1 and m2 are the mass of the object and R is the radius.
From the question above, we know that
R2 = 3R1
By substituting the following parameters, we get
F2/F1 = G.m1.m2 / R2² / G.m1.m2 / R1²
F2/F1 = G.m1.m2 / (3R1)² / G.m1.m2 / R1²
F2/F1 = 1/9
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Answer explanation !!!!!!!!!!
Answer:
65 meters for 5 seconds at a time
Explanation:
find the difference between the moon’s gravitational force on a 1 kg mass on the side of the earth facing the moon and an identical mass on the side of the earth opposite the moon (this difference is called the tidal force).
The tidal force is a gravitational phenomenon that, as a result of a gradient in the gravitational field from the other body, stretches a body along a line towards the center of mass of another body.
The surface gravity of the Moon is around 1/6th as strong, or 1.6 meters per second every second. Because the Moon is significantly less massive than Earth, its surface gravity is weaker. The surface gravity of a body is inversely related to the square of its radius but directly proportional to its mass.
The Earth is forced to orbit around the Earth-Moon center of mass, which is situated roughly 1600 km beneath the surface of the planet along a line connecting it to the Moon. It is possible to think of the tidal force as the difference between the force at the Earth's center and all other locations.
At sea level, where we can see the ocean tides, this distinction is evident. (Remember that the tidal forces' impact on sea level is measured relative to the baseline sea level.
As we previously observed, the rotation of the Earth causes a significant bulge along its equator. Here, we merely take into account the much smaller tidal bulge measured from that baseline sea level, which establishes the baseline sea level.)
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URGENT!! ILL GIVE
BRAINLIEST! AND 100 POINTS
Determine whether each statement describes an atom, a molecule, or both
Answer:
1.A
2.A
3.A
4.A
5.C
6.B
Explanation:
that's
your friend is catching a falling basketball after it has passed through the basket. her hands move straight down while catching the ball. it takes about 0.10 ss for the player to lower her hands to stop the ball. assume the mass of the ball is 0.60 kgkg, and that the ball has fallen a vertical distance of 1.2 mm before reaching the player's hand.
Average force that her hands exert on the ball while catching basketball whose mass is 600g = 29N
Given
clock t=0.1 s
ball mass m=0.6 kilogram
The previous distance traveled (measured in meters) was 1.2.
Net Force Exerted = 29N
Total Force used:
F= △P/△t
F= m△u/△t
F= m√2gd/△t
F= 0.6.√2.9.8.1.2/0.1
F= 0.6.4.849/0.1
F= 2.90/0.1
Force = 29N
What does force mean?
Power is an effect that can physically change the movement of an object. When an object with mass changes speed, it can be accelerated by a force. B. When leaving a stationary state. The obvious way to describe force is pushing or pulling. Force is a vector quantity because it has both magnitude and direction. It is calculated using Newton SI units (N). Force is represented by the letter F (formerly P).
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The question was incomplete. Check below the full question.
Your friend is catching a falling basketball after it has passed through the basket. Her hands move straight down while catching the ball. It takes about 0.10 s for the player to lower her hands to stop the ball. Assume the mass of the ball is 0.60 kg, and that the ball has fallen a vertical distance of 1.2 m before reaching the player's hand. Determine the average force that her hands exert on the ball while catching it.
5. How long would it take a truck to increase its speed from 10 m/s to 30 m/s if it does so
with uniform acceleration over a distance of 80 m ?
a) 8.0 s
b) 5.0 s
c) 4.0 s
d) 2.0 s
Answer:
C
Explanation:
Average velocity = (10+30) / 2 = 20 m/s
80 m / 20 m/s = 4 seconds
It would take a truck to increase its speed from 10 m/s to 30 m/s in 2.0 s. Option D is the correct answer.
To determine how long it would take for the truck to increase its speed from 10 m/s to 30 m/s, we can use the equations of motion. Option D is the correct answer.
The first equation we will use is: v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given:
Initial velocity, u = 10 m/s
Final velocity, v = 30 m/s
We need to find the time, t. Since the truck is accelerating uniformly, we can assume a constant acceleration. Next, we use another equation of motion: v² = u² + 2as, where s is the distance traveled.
Given:
Distance, s = 80 m
Initial velocity, u = 10 m/s
Final velocity, v = 30 m/s
Rearranging the equation, we get: s = (v² - u²) ÷ (2a)
Substituting the given values, we have: 80 = (30² - 10²) ÷ (2a)
Simplifying the equation, we find:
80 = (900 - 100) / (2a)
80 = 800 / (2a)
2a = 800 / 80
a = 10 m/s²
Now, we can go back to the first equation: v = u + at
Substituting the known values: 30 = 10 + 10t
Simplifying, we find: 10t = 20
t = 2 seconds
Therefore, it would take the truck 2.0 seconds to increase its speed from 10 m/s to 30 m/s.
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pls answer the 15a answer i cant understand it
Materials required for the experiment of limiting force borne by string:-
String balanceweightslight stringsweight hanger pan for spring balanceSandSteps of procedure for for the experiment of limiting force borne by string:-
First we have to tie a light string to the fixed support and then tie the other end with the weight hanger consists of weight.Add additional weight to the hanger again and again. And continue the same until the string is broken.Note down the weight (x) where the string is broken.Suspend spring balance to a support.Tie the light string at the end of the balance and at the other end suspend the pan for spring balance.Now place the weights (x-100 grams) in pan.Observe the reading in the spring balance.Add a small amount of sand in the pan by observing the readings.same is to be done till the string is broken.Learn more about limiting force here:- https://brainly.com/question/11371672
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