Answer:
We know that the force pulling the box in the positive x direction has a magnitude of m g sin 30 . Using Newtons Second Law, F = ma , we just need to solve for a :
ma=mgsin30
a=gsin30
=(10m/s2)(0.500)
=5m/s2
Answer:
A force is applied acting to the right. Assume that friction is negligible. Ikg 0.5N. For each question, one or more features of the system.
Explanation:hope this helps
a
(2) A 800 g block is pushed up an inclined plane (angled at 18°) with a velocity of 11.8 m/s. The first block slides up the
incline a distance of 2.2 m and strikes a second block with a mass of 300 g also moving at 3.4 m/s up the incline.
The two blocks hit and stick together. Determine the following:
(i) The maximum vertical height of the two blocks when they stop.
(ii) The time needed for the blocks to reach the bottom of the incline after the moment of impact.
(u = 0.19)
this is my attachment answer hope it's helpful to you
A long-distance runner runs at a constant speed of 4.8 m/s. How long does it take them to run 1.5 km?
Convert km to meters:
1km = 1000 m
1.5 km x 1000 = 1500 m
1500m / 4.8 m/s = 312.5 seconds
312.5 seconds / 60 seconds per minute =5.2 minutes = 5 minutes 12.5 seconds
More potential energy can be stored by moving against the magnetic force closer to a magnet?
Answer:
if your saying can it? then yes or if you are asking what type of magnetic force then its fr its actual self magnetic force
Explanation:
cecily is inflating her bicyble tyre with the pump below. when she pushes the plunger down, it is moving against a force appliefd by the air inside the cynlinder. this means that the plunger is doing ___
Answer:
"work against the force of the air in the tire"
The air in the tire provides a force opposing the force of the air provided by the plunger.
The length of a road is 35.5cm what will be the length in meters?
안녕하세요
answer is 355 meters ...
Answer:
0.355m
Explanation:
To convert from cm to m, divide by 100
35.5÷100=0.355m
please help me please help
Answer:
Answer → Distance is 5 × 10⁴ km
Explanation:
Gravitational formular
[tex]{ \tt{F = \frac{GMm}{ {r}^{2} } }} \\ [/tex]
F is the gravitational forceG is the universal gravitational constantr is the separation distanceM & m are the massesFor the first case;
[tex]{ \tt{F _{1} = \frac{GMm}{2.5 \times {10}^{4} } }} \\ [/tex]
For the second case;
[tex]{ \tt{F _{2} = \frac{GMm}{r _{2}} }} \\ [/tex]
but F2 = ½F1
Therefore, F1 = 2F2
Hence:
[tex]{ \tt{ \frac{GMm}{2.5 \times {10}^{4} } =2 \times \frac{GMm}{r _{2} } }} \\ \\ { \tt{ \frac{1}{2.5 \times {10}^{4} } = \frac{2}{r _{2} } }} \\ \\ { \tt{r _{2} = 2 \times 2.5 \times {10}^{4} }} \\ \\ { \underline{ \tt{ \: \: r _{2} = 5 \times {10}^{4} \: km \: \: }}}[/tex]
Answer: Distance is 5 × 10⁴ km
Explanation:
How can speed be defined
A) acceleration
B) change in velocity/time
C) distance / time
D) displacement/time
Answer:
Speed can be defined in (C) distance / time.
I hope this helped at all.
Kichelle enjoys balloon animals from the carnival. She just received a balloon giraffe that has an initial temperature of 39.0°C and a volume of 1.28 L. If Preston plays a trick on Kichelle, and puts her balloon giraffe into the freezer, what would be the new volume of the balloon if the temperature drops down to 8.0°C?
39°C=1.28L
8°C=? Less
=8°C/39°C×1.28L
=8/39×1.28L
=10.24/39
=512/195
=2.6L
ignore this i put the wrong grade sorry
Answer:
Oh its fine i am not in college but its still fine
Explanation:
Find the number of moles in a 28.0g sample of NH3.
PLS URGENTLY
Answer:
Molar mass of NH3=14+3=17g
[tex]mole = \frac{given \: mass}{molar \: mass} \\ mole = \frac{28}{17} \\ mole = 1.65[/tex]
BRAINLIST
What is the total amount of potential and kinetic energy in a system? *
1 point
A: Electrical energy
B: Heat energy
C: Mechanical energy
D: Nuclear energy
Answer:
C. Mechanical Energy
Explanation:
The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.
1. A stone is thrown vertically upward with velocity of 15 m/s at the same time, 10 m vertically above a second stone is allowed to fall. After what time and at what height do they collide (take g = 10 ms^-2)
Let the ball at the ground be A and that at the top be B.
Assume that you see the motion of A while sitting at B.
You(B) are obviously at rest w.r.t. yourselves. However you(B) and A have the same acceleration in the same direction. Thus acceleration of A w.r.t. you(B) is 0.
Thus, to you, it will appear as if A is travelling towards you with an uniform speed of 20m/s.
Dividing the distance 20m with that speed, we get that A reaches you(B) in 1s.
Now, you have considered these motions and time by assuming you were stationed at B. However, the time taken for A to meet B shouldn't be dependent on which reference frame you assume. Thus, time taken for A and B to collide is 1s.
Since B falls freely, he covers a distance of 1/2(g)(1²)=5 (assuming g = 10m/s²)
Thus, they meet at a height of (20–5)m = 15m from the ground.
a helicopter cruises at a constant velocity when the total thrust of the engines is 5,000 N. How much air resistance acts on the jet ??
Hi there!
[tex]\large\boxed{F_{A} = -5000N}}[/tex]
For an object to be moving at constant velocity:
∑F = 0 (The sum of the forces acting on the object MUST be 0 N)
We can do a summation of forces for this helicopter in the HORIZONTAL direction:
∑Fₓ = Ft (Force of Thrust) + Fa (Force of air resistance) = 0
Thus, if we substitute in the given value for Ft:
Ft = -Fa
5000N = -Fa
Fa = -5000N
A swimming pool has dimension of 30m×10m×3m. When it is filled with water , what is the thrust on the bottom and sides
Answer:
3,900m³
Explanation:
A runner slows down after completing a race. Her deceleration is 0.25 m/s2. After 5 s she is travelling at 4 m/s, determine her initial velocity.
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
The initial velocity of the runner is 5.25 m/s.
What is deceleration?
You've probably seen that when there's a lot of traffic and there are more bikes blocking us, we tend to slow down the speed of our bikes.
Therefore, deceleration is defined as a reduction in speed as the body moves away from the beginning location. The opposite of acceleration is deceleration.
Given parameters:
Deceleration of the runner: a = 0.25 m/s².
Time interval: t = 5 second.
Final velocity: v = 4 m/s.
We have to find initial velocity of the runner: u = ?
From definition of deceleration, it can be written that:
deceleration = (initial velocity - final velocity) time interval
0.25 = (u -4)/5
u = 5.25 m/s.
Hence, initial velocity of the runner is 5.25 m/s.
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The diagram shows what happened in the Rutherford and Marsden scattering experiment. Complete this sentence: The particle shown in red will come straight back from the foil because it is __________ by the charge in the gold nucleus.
From Rutherford and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.
Rutherford proposed Planetary model atom, which visualized an atom to consists of a positively charged heavy core called the nucleus around which negatively charged electrons circle in orbits much as planets move round the sun
Thus, from Rutherford and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.
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the answer is reflected or u can say repelled
name the substance that combines with water in photosynthesis
Answer:
carbon dioxide
Explanation:
its common sense
one end of a light inextensible string is attached to a tool box of mass 2.5 kg which is lying on a horizontal table. The string passing over a smooth pulley and is tied at the other end to a bag of mass 1.4kg as shown in the diagram
a. If the toolbox on the point of sliding, find the value for μ, the coefficient of friction.
b. Supposing the coefficient of friction between the tool boxe and the table is 0.20, calculate the acceleration of the system and the tension in the string gravity as 10 m per second square
Answer:
the answer is A
Explanation:
1 point
What is the potential energy of a bird flying, if the bird has 1000 J of total
energy and 450 J of kinetic energy? (only put the number, no units or
commas)
Answer:
the bird has 550J of potential energy
Explanation:
PEtotal=PE+KE
1000J=PE+450J
subtracting the kinetic energy from the total we get:
1000J-450J=PE
550J=PE
Which explains why scientists often prefer to measure mass rather than weight?
It is always the same.
It is affected by gravity.
Answer:
scientists often prefer to measure mass rather than weight because weight is affected by gravity while mass is not. Mass is always the same.
Explanation:
Weight is a force rather than a measure of the amount of matter something contains. The same amount of matter can have different weights depending on where it is. Mass directly measures the amount of matter present and remains the same no matter where it's measured.
what is the purpose of delivering medical aid through drones
Find the gravitational potential energy of a body of mass 25kg,kept at a height of 4m If g=10m/s'.
Answer:
1000 JExplanation:
The gravitational potential energy of a body can be found by using the formula
GPE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 10 m/s²
From the question we have
GPE = 25 × 10 × 4 = 1000
We have the final answer as
1000 JHope this helps you
2. Un niño hace girar con la mano una pelota de hule que se encuentra sujeta mediante un cordón de
0.75 m de longitud. Si la pelota da 0.7 vueltas a cada segundo, entonces,
a) ¿Cuál es el periodo de la pelota?
b) ¿Cuál es la velocidad lineal de la pelota?
Cuando el niño hace girar 0.7 vueltas por segundo una pelota de hule que se encuentra sujeta mediante un cordón de 0.75 m de longitud, tenemos que:
a) El periodo de la pelota es 1.43 segundos.
b) La velocidad lineal de la pelota es 3.3 m/s.
a) El periodo de la pelota está dado por:
[tex] T = \frac{2\pi}{\omega} [/tex]
En donde:
ω: es la velocidad angular
Dado que la pelota da 0.7 vueltas (revoluciones) cada segundo, la velocidad angular es:
[tex] \omega = \frac{0.7 \:rev}{s}*\frac{2\pi rad}{1 \:rev} = 4.40 rad/s [/tex]
Entonces, el periodo es:
[tex]T = \frac{2\pi}{\omega} = \frac{2\pi}{4.40 rad/s} = 1.43 s[/tex]
b) La velocidad lineal de la pelota se puede calcular usando la siguiente ecuación:
[tex] v = \omega r [/tex]
En donde:
r: es el radio de la circunferencia = longitud del cordón = 0.75 m
[tex]v = \omega r =4.40 rad/s*0.75 m = 3.3 m/s[/tex]
Por lo tanto, la velocidad lineal de la pelota es 3.3 m/s.
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When there is a temperature inversion, you would expect to experience Group of answer choices clouds with extensive vertical development above an inversion aloft. good visibility in the lower levels of the atmosphere and poor visibility above an inversion aloft. an increase in temperature as altitude increases.
Temperature inversion leads to an increase in temperature as altitude increases.
The term temperature inversion refers to a situation in which a layer of warm air lies over a layer of cool air. This is also referred to as thermal inversion. This occurs when the air below to loose heat rapidly.
One of the effects of temperature inversion is reduction in visibility. So, thermal inversion leads to an increase in temperature as altitude increases.
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Explain the relationship between the frequency, energy, and wavelength of a wave.
Answer:
Explanation:
The energy equation is E=hv
a soccer ball is kicked at an angle of 35° and it lands on even ground.
A) What angle will produce the same range?
B) Which angle of the two will produce the highest ball?
C) Which angle of the two will produce the ball that is in the air the longest?
D) What angle in this situation would produce the furthest range?
Hi there!
A)
The angle that will produce the same range is the compliment of 35°.
Thus, kicking the ball at 55° will result in the same range.
We can prove this by using the derived range equation:
[tex]R = \frac{v^2sin2\theta}{g}[/tex]
An angle of 35° yields:
[tex]R = \frac{v^2sin(2*35)}{g} = .939R[/tex]
An angle of 55° yields:
[tex]R = \frac{v^2sin(2*55)}{g} = .939R[/tex]
Both are the same, thus indicating that 55° produces the same range.
B)
The angle of 55° will produce the higher ball because the VERTICAL component of the ball's velocity is greater compared to kicking the ball at 35° degree.
sin(55) > sin(35)
C)
The angle of 55° will result in the ball being in the air the longest because when a ball is in the air (assuming no air resistance), the ball experiences an acceleration due to gravity of -9.8 m/s², causing the vertical velocity to decrease until it eventually reaches 0 m/s at the top of its path. A greater initial vertical velocity means that it will take longer for the ball to fall.
We can prove this using:
vf = vi + at
0 = vy - 9.8t
vy/9.8 = t
Greater vy (vertical component of velocity) ⇒ greater time taken.
D)
The angle that would result in the furthest range is 45°.
We can prove this using calculus. Recall the above range equation:
[tex]R = \frac{v^2sin2\theta}{g}[/tex]
We can take the derivative and use the first-derivative test to find its critical point:
[tex]\frac{dR}{d\theta} = \frac{v^22cos2\theta}{g} = 0[/tex]
Evaluate:
[tex]v^22cos2\theta = 0 \\\\cos2\theta = 0 \\\\2\theta = 90^o\\\\\boxed{\theta = 45^o}[/tex]
A. The angle that will produce the same range as a 35° kick is 53°.
B. The angle that produces the highest ball is 45°.
C. The angle that produces the ball that is in the air the longest is 45°.
D. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.
A. The range of a projectile is the horizontal distance it travels before it hits the ground. The range of a projectile is determined by the initial velocity, the angle of projection, and the acceleration due to gravity.
For a given initial velocity, the range of a projectile is maximized when the angle of projection is 45°. However, if the ground is not level, the range of a projectile can be maximized at other angles.
In the case of a soccer ball kicked at an angle of 35°, the range will be maximized at an angle of 53°. This is because the range of a projectile is maximized when the vertical component of the initial velocity is equal to the horizontal component of the initial velocity.
The angle of 53° is the angle that produces a vertical component of the initial velocity that is equal to the horizontal component of the initial velocity when the ball is kicked at an angle of 35°.
Therefore, the angle that will produce the same range as a 35° kick is 53°.
B. The angle that produces the highest ball is 45°.
C. The angle that produces the ball that is in the air the longest is 45°.
D. The angle that produces the furthest range depends on the initial velocity and the acceleration due to gravity. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.
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What is the potential energy of a 50kg car on top of a 600m hill?
Answer:
294,000 JExplanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
From the question we have
PE = 50 × 9.8 × 600 = 294,000
We have the final answer as
294,000 JHope this helps you
A 60 kg gorilla named Anthony Falcon is standing on his skateboard. This is on planet Erf,
where a = g = -10m/s2
If Anthony Falcon has a weight of 600 N, with what amount of force is their body pushing
down on the skateboard with?
Two cars A and B are moving along a straight road in the OPPOSITE direction with velocities of 25 km/h and 40 km/h, respectively. Find the velocity of car B relative to car
Answer:
65km/h
Explanation:
Here,
Relative velocity of car moving in different directions with different velocities =Va + Vb
=25+40
=65 km/h
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
A small ball is attached to one end of a spring that has an unstrained length of 0.201 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.41 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0176 m.
Required:
By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?
The extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.
The given parameters;
unstrained length, l₁ = 0.201 mextension of the string, x = 0.0176 mspeed of the ball, v = 3.41 m/sThe radius of the circular path when spring is stretched is calculated as;
R = l₁ + x
R = 0.201 + 0.0176
R = 0.2186 m
The spring constant is calculated as follows;
[tex]F = ma\\\\Kx = \frac{mv^2}{R} \\\\K = \frac{mv^2}{Rx} \\\\K = \frac{(3.41)^2 m}{0.2186 \times 0.0176} \\\\K = 3,022.4 m \ N/m[/tex]
The extension of the spring when the ball is allowed to hang straight down, motionless;
[tex]F = mg\\\\Kx=mg\\\\x = \frac{mg}{K} \\\\x = \frac{9.8 m}{3022.4 m} \\\\x = 0.0032 \ m[/tex]
Thus, the extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.
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