Answer:
[tex]47 \ \frac{m}{s}[/tex]
Explanation:
s = displacement (m)
u = initial velocity [tex](\frac{m}{s})[/tex]
v = final velocity [tex](\frac{m}{s})[/tex]
a = acceleration [tex](\frac{m}{s^{2} })[/tex]
t = time (s)
s = 235
a = -4.7
v = 0
v² = u² + 2as
(0)² = u² + 2(-4.7)(235)
u² - 2209 = 0
u² = 2209
u = 47
Answer:
[tex]\boxed {\boxed {\sf 47 \ m/s}}[/tex]
Explanation:
We are asked to find the initial velocity of the car before it began to accelerate.
We are given the acceleration, distance, and final velocity, so we will use the following kinematic equation:
[tex]{v_f}^2 = {v_i}^2 + 2ad[/tex]
The car's acceleration is -4.7 meters per second square. It traveled a distance of 235 meters. It came to rest, or a final velocity of 0 meters per second.
a= -4.7 m/s²d= 235 m [tex]v_f[/tex]= 0 m/sSubstitute the values into the formula.
[tex](0 \ m/s)^2 = {v_i}^2 + 2 (-4.7 \ m/s^2)(235 \ m)[/tex]
[tex]0 = {v_i}^2 + 2 (-4.7 \ m/s^2)(235 \ m)[/tex]
Multiply the numbers in parentheses.
[tex]0= {v_i}^2 + (-2209 \ m^2 / s^2)[/tex]
Add -2209 to both sides of the equation.
[tex]0+ 2209 \ m^2 /s^2 = {v_i}^2+ ( -2209 \ m^2 /s^2 )+ 2209 \ m^2 /s^2[/tex]
[tex]2209 \ m^2 /s^2 = {v_i}^2[/tex]
Take the square root of both sides.
[tex]\sqrt {2209 \ m^2 /s^2} = \sqrt {{v_i}^2[/tex]
[tex]\sqrt {2209 \ m^2 /s^2} = v_i[/tex]
[tex]47 \ m/s = v_i[/tex]
The inital velocity of the car was 47 meters per second.
A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?
Answer:
Explanation:
The work increased the potential energy
W = PE = mgh = 40(9.8)(15) = 5880 J(oules)
Which statement best describes wavelength?
wavelength is the speed at which a wave travels
wavelength is the height vertically of a wave
wavelength is the length of wave per sec
wavelength is the horizontal length of one complete wave cycle
Answer:
B
Explanation:
Wave length is the height perpendicular verically of a wave
Answer:
b
Explanation:
makes the most sense to me
8 N to the left , and 4 N to the right. Find the net force. Is this balanced?
Explanation:
12N by first law of newton is net force after colloision
Tony brought 9 2/3pitchers of juice to a volleyball game, and the players drank3 7/8pitchers of it. How much juice is left?
Rewrite the amounts as improper fractions:
9 2/3 = 29/3
3 7/8 = 31/8
Rewrite both fractions with a common denominator
29/3 = 232/24
31/8 = 93/24
Now subtract: 232/24 - 93/24 = 139/24
Rewrite as a proper fraction: 5 19/24
Answer 5 19/24
An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)
0.8 seconds
Explanation:
time of flight = 2u/g
u=4m/s
g=10
= 8/10
= 0.8 sec
just a trial...not sure!!!
Given :
∅ = 60⁰
u = 4 m/s
g = 10m/s²
to find :
T = ?
Solution :
as per formula,
[tex]t = \frac{2u \: sin \theta}{g} [/tex]
now put the value : [tex]t \: = \frac{2 \times 4 \times sin \: 60}{10} [/tex]
as we know [tex] sin60 \: = \frac{ \sqrt{3} }{2} [/tex]
therefore,
[tex]t \: = \frac{8 \times \frac{ \sqrt{3} }{2} }{10} [/tex]
as we solve this we get,
[tex]t \: = \frac{ 2\sqrt{3} }{5} [/tex]
that's t = 0.69 sec
[tex]\sf\fbox\red{\:I \:hope \:it's \:helpful \:to \:you}[/tex]
A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 650 m/s . Part A What is the recoil speed of the hunter
Answer:
Explanation:
momentum is conserved. Initial momentum was zero, so final total momentum must also be zero
0.042(650) + 70v = 0
v = -0.39 m/s
|v| = 0.39 m/s
The force shown in the figure(Figure 1) moves an object from x = 0 to x = 0.75 m.
1/How much work is done by the force?
2/How much work is done by the force if the object moves from x = 0.20 m to x = 0.55 m ?
Answer:
(a) The force changes its magnitude with respect to displacement, hence the total work will be sum of increment of work in three steps:-
step 1 . from 0 to 0.25m .
force = 0.6 N
displacement= 0.25m
work done =( force × displacement) = (0.25 × 0.6 ) = 0.15 joule.
step 2:- .
work done in moving from 0.25 to 0.50 m.
work done = ( force × displacement) = ( 0.4 × 0.25) = 0.10 Joule. .
step 3 :-
work done in moving from 0.50 to 0.75 m
work done = 0.8 × 0.25 = 0.200 joule.
hence total work done = ( 0.20+0.10+0.15) = 0.45 joule. ans
(b) similar concept you have to use here also.
step 1:
from 0.20 to 0.50, force of magnitude 0.4 N acts on the object.
Work done = ( 0.50-0.20)× 0.4 = 0.30 × 0.4 = 0.12 joule.
step :- 2
from 0.50 to 0.55 , force of magnitude 0.8 N acts on the block.
work done = 0.8× ( 0.55- 0.50) = 0.04 joule
total work done = 0.04 + 0.12 = 0.16 joule. ans
Four balls with different masses are dropped from the heights shown. Air resistance may be ignored. Which ball has the greatest average speed?
Answer:
The one falling from the greatest height will have the greatest speed.
h = 1/2 g t^2 time for ball to fall distance h
h2 / h1 = t2^2 / t1^2 dividing equations
h2 / t2^2 = h1 / t1^2
Let v be the average speed (v2 = h2 / t2)
1 / t2 * v2 = 1 / t1 * v1
v2 / v1 = t2 / t1 the one taking the longest to fall has the greater av. speed
Check:
h4 / h1 = t4^2 / t1^2 or
t4 / t1 = (h4 / h1)^1/2
In this case t4 / t1 = (4 / 1)^1/2 = 2 or twice the average speed
t1 = (2 h / g)^1/2 = .2^1/2 = .447 using g = 10
t4 = (2 h / g)^1/2 = .8^1/2 = .894
v1 = 1 / .447 = 2.24 m/s average speed
v4 = 4 / .894 = 4.47 or twice the average speed
A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly
released and slides to rest in 5 sec. What is the magnitude of the frictional force
producing this deceleration?
Answer:
The frictional force producing this deceleration would have a magnitude of [tex]4\; \rm N[/tex].
Explanation:
The velocity of this object changed by [tex]\Delta v = (-10\; \rm m\cdot s^{-1})[/tex] in [tex]\Delta t = 5\; \rm s[/tex]. The acceleration of this object would be:
[tex]\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:
[tex]\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}[/tex].
([tex]1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}[/tex].)
If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be [tex]4\; {\rm N}[/tex], same as the magnitude of the net force on this object.
Use the sentence to answer the question.
Light is affected by gravity.
Which inference can be made based on this fact?
(1 point)
Light behaves differently in space than on Earth.
Light behaves differently in space than on Earth.
Gravity causes light to refract.
Gravity causes light to refract.
Light moves faster in space than on Earth.
Light moves faster in space than on Earth.
Stronger gravity causes an increase in light.
Answer:
Light behaves differently in space than on Earth.
Explanation:
Because the gravity field is greater near earth than in most of space. Not the areas near stars, black holes, pulsars, and such but in the vast emptyness between the clumpy spots.
What is the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25
s with 7,500 N of force?
128 kg
3991 kg
0.017 kg
14,091 kg
The mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.
HOW TO CALCULATE MASS:
The mass of an object can be calculated by dividing the force applied to the object by its acceleration. According to this question, a bus can accelerate from rest to 15.5 m/s over 8.25s. The acceleration can be calculated as follows:a = (v - u)/ta = 15.5 - 0/8.25a = 15.5/8.25a = 1.88m/s²The mass of the bus = 7500N ÷ 1.88m/s²The mass of the bus = 3989.4kgTherefore, the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.Learn more about mass at: https://brainly.com/question/20259048?referrer=searchResults
write 2 situations in which the energy changes mentioned occur
Answer:
The types of energy is bond breaking and bond forming in chemical energy.
Explanation:
During Chemical reaction energy is required either for breaking up bonds in case of reactants and building bonds to form products.
The chemical reaction in which energy is released is called exothermic reactions, which is released due to making up the bonds.
The chemical reaction in which energy is absorbed is called endothermic reactions, in which energy is absorbed for breaking up the bonds.
Qué velocidad –en m/s– tiene un móvil, que recorre 15 km en 20 minutos
(es para hoy por faaaa)
Answer:
Explanation:
15 km(1000 m / km) / (20 min(60 s/min)) = 12.5 m/s
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?
Answer:
Explanation:
T = 2π[tex]\sqrt{L/g}[/tex]
(T / 2π)² = L/g
g = 4π²L/T²
g = 4π²(0.75000)/(1.7357)²
g = 9.82814766...
g = 9.8281 m/s²
Physics!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
What is the formula for calculating distance?
QA: Speed x Time -- Speed/Time -- Time/Speed
Answer:
x=v.t
The answer: Distance= Speed x Time
And also
Time = Distance/Speed
Speed= Distance/Time
K
Mission CG9: Weightlessness
Consider the several locations along a roller coaster
track. In which location(s) would the riders feel less
than their normal weight? Select all that apply.
Location A
Location B
Location C
a
=-10 m/s/s, dn
--2 m/s/s, up
a--6 m/s/s, dn
Location D
Location E
x=-12 m/s/s, dn
---6 m/s/s, up
The locations where the riders feel less than their normal weight are Location A, Location C and Location D.
The given parameters;
Location A, a = 10 m/s² downLocation B, a = 2 m/s² upLocation C, a = 6 m/s² downLocation D, a = 12 m/s² downLocation E, a = 6 m/s² upThe normal weight of the riders is calculated by applying Newton's second law of motion as follows;
W = mg
W = 9.8m
The apparent weight of the riders for the upward acceleration is calculated as follows;
[tex]R = m(g + a)[/tex]
The apparent weight of the riders for the downward acceleration is calculated as follows;
[tex]R = m(g - a)[/tex]
The apparent weight of the riders at location A is calculated as follows;
[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]
The apparent weight of the riders at location B is calculated as follows;
[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]
The apparent weight of the riders at location C is calculated as follows;
[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]
The apparent weight of the riders at location D is calculated as follows;
[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]
The apparent weight of the riders at location E is calculated as follows;
[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]
Thus, the locations where the riders feel less than their normal weight are;
Location ALocation CLocation D.Learn more about Newton's second law and reading of a scale here: https://brainly.com/question/11603452
A CD has a diameter of 12.0 cm. If the CD is rotating at a constant angular speed of 200 revolutions per minute, then the tangential velocity of a point on the circumference is:
Hi there!
Converting from angular velocity to tangential velocity can be done by:
v = ωr
v = tangential velocity (m/s)
ω = angular velocity (rad/sec)
r = radius (m)
Convert 12 cm to meters:
100 cm = 1 m
12 cm = 0.12 m
Now, convert rev/min to rad/sec:
[tex]{\frac{200rev}{min}} * \frac{1min}{60s} * \frac{2\pi rad}{1 rev} = 20.94 rad/sec[/tex]
v = 20.94 · 0.12 = 2.51 m/s
Which of the following is an incorrect statement?
The pulley is a special type of wheel and axle.
The mechanical advantage of a pulley with three ropes is one.
The mechanical advantage of a block and tackle pulley with two pulleys is two.
The mechanical advantage of a fixed pulley is one.
PLZ HELP WILL MARK BRAINLIEST
Answer:
Explanation:
The mechanical advantage of a pulley with three ropes is one.
is incorrect. It is 3
A .223 rifle bullet, traveling at 370.m/s, hits a block of pine wood, and goes in, to a depth of 0.130m. The mass of the bullet is 1.75g (mass = 0.00175kg). Assume a constant slowing force. (A) What is the acceleration that the bullet experiences as it comes to a stop? (B) How much time is required for the bullet to stop? (C) What force, in Newtons, does the wood exert on the bullet?
Answer:
Explanation:
A)
v² = u² + 2as
a = (v² - u²) / 2s
a = (0² - 370²) / (2(0.130))
a = -526,538 m/s²
B)
t = v/a
t = 370 / 526538
t = 0.0007027... s or 0.7 ms
C)
F = ma
F = 0.00175(526,538) = 921.442307... = 921 N
Which of these is Newton's 3rd law of motion?
Equal and opposite forces
F = m x a
Inertia
Gravity
Chronic diseases are also know as
Answer:
Chronic Condition or Long-term Illness
Explanation:
They can also be known as Chronic Conditions or Long-Term Illnesses, hope this helps.
Answer:
Chronic condition, also called long-term condition.
Explanation:
You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weight from different heights and record the time it takes for the object to hit the ground when dropped from different heights. The data you collect is shown below:
Height dropped (m)
Time to fall (s)
1.0
0.45
2.0
0.63
3.0
0.78
4.0
0.89
Next, you plan to drop a 5 kg weight from the same heights. How will your time values in your new data table for the 5 kg weight compare to the time values in the old data table for the 1 kg weight?
The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
The given parameters;
Mass of the first object, m1 = 1 kg
Mass of the second object, m2 = 5 kg
The final velocity of the objects during the downward motion is calculated as follows;
[tex]v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt[/tex]
The time of motion of the object from the given height is calculated as;
[tex]h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }[/tex]
The time of motion of each object is independent of mass of the object.
Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.
Learn more about time of motion here: https://brainly.com/question/2364404
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula
KE = one half.mv2, where m is the mass and v is the speed (velocity). Assume that the speed of the soda bottle falling from a height of
0.8 m will be 4 m/s, and use this speed for each calculation.
Record your calculations in Table A of your Student Guide.
When the mass of the bottle is 0.125 kg, the KE is
✔ 1
kg m2/s2.
When the mass of the bottle is 0.250 kg, the KE is
✔ 2
kg m2/s2.
When the mass of the bottle is 0.375 kg, the KE is
✔ 3
kg m2/s2.
When the mass of the bottle is 0.500 kg, the KE is
✔ 4
kg m2/s2.
Answer included in Question
Answer:
kinetic energy is given as KE = (0.5) m v²given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg KE = (0.5) m v² = (0.5) (0.125) (4)²
Explanation:
Answer:
1. 0.5 2. 2 3. 3.75 4. 5
Explanation:
define parking orbit?
Answer:
An orbit of a spacecraft from which the spacecraft or another vehicle may be launched on a new trajectory.
The diagram shows the velocity-time graph for a car travelling in a straight line along a road. Calculate the acceleration between t = 2.0 s and t = 5.0 s.
Answer:
a = Δv/Δt = (0 - 20) / (5 - 2) = -6⅔ m/s²
Which object would have more momentum?
A 2 kg ball rolling at 4 m/s
A 5 kg ball rolling at 2 m/s
(Hint: Use momentum = mass * velocity)
a
Not enough information to determine which ball has more momentum.
b
They have the same momentum.
c
The 2 kg ball will have more momentum.
d
The 5 kg ball will have more momentum.
Answer:
b is the ans......
Explanation:
"b" (and any subsequent words) was ignored because we limit queries to 32 words.
2. An object is falling under gravity with terminal velocity. What is happening to its speed?
A. It is decreasing to a lower value.
B. It is increasing
C. It is decreasing to zero.
D. It is staying constant.
Answer:
the speed of the object has become constant.
Explanation:
At terminal velocity, air resistance equals in magnitude the weight of the falling object. Because the two are oppositely directed forces, the total force on the object is zero, and the speed of the object has become constant.
Which of these do not affect fluid friction?
The surface of an object
The viscosity (thickness) of fluid an object is in
The shape of an object
The weight of an object
2. A ray of light is incident at 60° in the air on an air glass plane surface find the angle of refraction in the glass. (mew for glass=1.5)
[tex] [/tex]
Answer:
35.2644
I suppose mew is refractive index
Explanation:
( sin i ) / (sin r) = refractive index
( sin 60) / (sin r) = 1.5
( sin 60) / 1.5 =sin r
r=35.844
sorry if I'm wrong
A racing car traveling with constant increases its speed from 10 m/s; 30 m/s over a distance of 60 mlong does this take? to
Answer:
Explanation:
constant acceleration???
assume it to be so
average speed is (10 + 30) / 2 = 20 m/s
t = d/v = 60/20 = 3 s