A 5 kg ball is dropped from a height of 200 meters on Earth.
a) What is the ball’s potential energy at its original height?

Answer:

638 m.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 94 m/s

Final velocity (v) = 22 m/s

Time (t) = 11 s

Distance (s) =?

We can obtain the distance travelled by using the following formula:

s = (u + v) t /2

s = (94 + 22) × 11 /2

s = 116 × 11 /2

s = 1276 /2

s = 638 m

Thus, the distance travelled is 638 m.

Answer:

F = 250 [N] (Force exerted by Sunita)

Explanation:

First we must understand the use of the lever, in the attached picture we see that it is fulcrum and other features.

The moment that the load exerts with respect to the fulcrum should be calculated, as the distance is 1 [m].

[tex]M=F*d\\M=1000*1\\M=1000[N*m][/tex]

Now Sunita must exert the same moment using the remaining distance (4 [m]).

[tex]1000=F_{sunita}*4\\F_{sunita}=1000/4\\F_{sunita}=250 [N][/tex]

Answer:

The average power developed by a motor is 4905 Watts.

Explanation:

The average power is given by:

[tex] P = \frac{W}{t} [/tex]

Where:

W: is the work

t: is the time = 8.0 s

First, let's find the work:

[tex] W = F*d [/tex]

Where:

F: is the force

d: is the displacement = 10.0 m

The force is in the vertical motion, and since the movement of the mass is at constant speed the force is:

[tex] F = mg = 400 kg*9.81 m/s^{2} = 3924 N [/tex]

Hence, the average power is:

[tex] P = \frac{F*d}{t} = \frac{3924 N*10.0 m}{8.0 s} = 4905 J/s = 4905 W [/tex]                    

Therefore, the average power developed by a motor is 4905 Watts.

I hope it helps you!                  

The average power developed by a motor as it lifts the mass is 4,900 W.

The given parameters:

The average power developed by a motor as it lifts the mass is calculated as follows;

[tex]P = \frac{E}{t} \\\\P = \frac{mgh}{t} \\\\P = \frac{400 \times 9.8 \times 10}{8} \\\\P = 4,900 \ J/s[/tex]

Thus, the average power developed by a motor as it lifts the mass is 4,900 W.

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Answer:

1.1x10^-4 N

Explanation:

F = G[(m_1*m_2)/r^2]

G = Gravitational constant 6.67433x10^-11 (N*m^2)/(kg^2)

m_1 = mass 1

m_2 = mass 2

r = radius between objects

F = G[(0.5kg*0.33kg)/(0.001m)^2]

F = 1.1x10^-4 N

Answer:

5m/s/s

Explanation:

The units for velocity is m/s.

The units for acceleration is m/s/s

So, Determine the change in velocity and divide it by the time that change took place.

(30m/s - 10m/s)/4s = 20m/s/(4s) = 5m/s/s

Answer:

Since it is falling freely, the only force on it is its weight, w. w = m ⋅ g = 1000kg ⋅ 9.8m s2 = 9800N To draw a Free Body Diagram, draw an elevator cage (I am sure you would get lots of points for drawing it with intricate detail) with a downward force of 9800 N. I hope this helps,

Explanation:

Explanation:

❀ [tex] \underline {{\underline{ \text{Given} }}}: [/tex]

❀ [tex] \underline{ \underline{ \text{To \: find}}} : [/tex]

❀ [tex] \underline{ \underline{ \text{Solution}}} : [/tex]

[tex] \boxed{ \sf{force = mass \times acceleration}}[/tex]

Plug the known values :

⟶[tex] \sf{1000 \times 9.8}[/tex]

⟶ [tex] \sf{9800 \: N}[/tex]

[tex] \red{\boxed{ \boxed{ \tt{⟿ \: Our \: final \: answer : 9800 \: N}}}}[/tex]

Hope I helped !♡

Have a wonderful day / night ! ツ

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Answer:

C

Explanation:

Energy is neither created or destroyed, but it can change forms...

When the object is at rest, there is a zero net force due the cancellation of the object's weight w with the normal force n of the table pushing up on the object, so that by Newton's second law,

∑ F = n - w = 0   →   n = w = mg = 112.5 N ≈ 113 N

where m = 12.5 kg and g = 9.80 m/s².

The minimum force F needed to overcome maximum static friction f and get the object moving is

F > f = 0.50 n = 61.25 N ≈ 61.3 N

which means a push of F = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.

So:

(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.

(b) Friction has a magnitude of 15 N because it balances the pushing force.

(c) The object is in equilibrium and not moving, so the acceleration is zero.

Answer:

yes

Explanation:

if your destination is on the top floor, but you are facing towards the bottom floor, you are facing the wrong way. It's all relative. Honestly, I hate this question, because then it turns into, is it possible to go backwards anywhere. Backwards and forwards, right and left, it's all relative to which way you are facing and where you want to go. If something is on the left and you turn 180 degrees, then that thing is on the right. If you are going forward towards your house and you turn around, you are now going backwards, relative to your house. But if your perspective is relative to that tree in the opposite direction of your house, you are now facing the right way. See, it's all relative to where you are going.

Answer:

Explanation:

From the given information; the diagram below shows a clearer understanding.

The blood pressure in the brain [tex]P_{brain} = P_{heart} - \delta ( a - g ) h[/tex]

= 13300 - 1060 (12-9.81) 0.35

=  13300 -  1060 (2.19) 0.35

= 13300 - 812.49

= 12487.51  Pa

The blood pressure in the feet [tex]P_{feet} = P_{heart} + \delta (a + g) h[/tex]

= 13300 + 1060 (12 + 9.81) 1.45

= 13300 + 1060( 21.81 ) 1.45

= 13300 + 33521.97

= 46821.97 Pa

Answer:

Explanation:

[tex]P_brain = P_heart - y(a - g)h\\\\P_brain = 13300 - 1060 (12-9.81)0.35\\\\P_brain = 12487.51 pa\\\\[/tex]

[tex]P_feet = P_heart + y(a+g)*h_r\\\\P_feet = 13300 + 1060(12+9.81)*(1.8-0.35)\\\\P_feet = 46821.97pa[/tex]

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Answer:

(a) [tex]Net\ Force = 2N[/tex] towards the right

(b) The green team wins

Explanation:

Given

[tex]Blue = 40N[/tex]

[tex]Green = 42N[/tex]

Solving (a): The net force

Because the force in the opposite directions, the net force is the subtraction of the smaller force from the bigger one.

So, we have:

[tex]Net\ Force = Green - Blue[/tex]

Substitute values for Blue and Green

[tex]Net\ Force = 42N- 40N[/tex]

[tex]Net\ Force = 2N[/tex]

Solving (b): Who wins the round

The body will move toward the Green Team because the force larger [tex](42N > 40N)[/tex]

So, the green team will win

Answer:

It is because there are a odd number of wavelengths fitting onto the string

Explanation:

The general formula for the harmonics on a string loose at one end is f = nv/2L where n = odd integer, v = speed of wave in string and L = length of string.

For a string loose at one end, there are a odd number of wavelengths which can fit onto the string. Since this is the case, we only have harmonics of odd numbers and thus we have even number harmonics missing.

Answer:

Power = 0.25 Watts

Explanation:

Given:

Work done = 0.50 Joule of work

Time taken = 2.0 second

Find:

Power

Computation:

Power = Work done / Time taken

Power = 0.50 J / 2

Power = 0.25 Watts

If the tired squirrel does all this work in 2 seconds with work of 0.5 J the power is 0.25 Watts.

Given:

Mass, m = 1 kg

Work, W = 0.5 J

Time, t = 2 s

The concept of power, which measures the pace at which work is done or the rate at which energy is transmitted or converted, is crucial to physics. It measures how quickly energy is produced or utilized in a system, to put it simply. In the International System of Units (SI), power is a scalar quantity and is measured in watts (W).

The power is given by:

Power  = Work ÷ Time

P = 0.50 Joules ÷ 2 seconds

P  = 0.25 Watts

Hence, the power exerted by the tired squirrel is 0.25 Watts.

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Explanation:

The mass of a truck = 5000 kg

Mass of a person = 50 kg

The momentum of a body is given by :

p = mv

m is mass

If momentum of the truck and the person is same. Let v₁ and v₂ are velocity of the truck and the person.

[tex]5000\times v_1=50\times v_2\\\\\dfrac{v_1}{v_2}=\dfrac{50}{5000}\\\\\dfrac{v_1}{v_2}=\dfrac{1}{100}\\\\v_2=100v_1[/tex]

So, it can only possible if the velocity of the person is 100 times of the velocity of the truck.

Answer:

.

Explanation:

Answer:

8.2N

Explanation:

I took the test and got it right :)

Answer:

The potential difference between the points is: 0.1 V

Explanation:

Given

To determine

The potential difference between the points is

We can use the formula to determine the potential difference between the points

[tex]V_A-V_B=\frac{W}{q_0}[/tex]

where W is done by moving charge q₀ from point A and point B

substituting W = 2J, and q₀ = 20 C

              [tex]=\frac{2}{20}[/tex]

               [tex]= 0.1[/tex] V

Therefore, the potential difference between the points is: 0.1 V

Answer:

your answer is C

Explanation:

Answer:

1.000 metre

Explanation:

I hope it's helpful

Answer:

Distance.  A light-year is the distance a single photon of light can travel in empty space in a year.

Answer:

D-Distance

Explanation:

Answer:

6.75 m/s²

Explanation:

The following data were obtained from the question:

Mass (m) of object = 40 Kg

Force applied (Fₐ) = 300 N

Force of friction (Fբ) = 30 N

Acceleration (a) =?

Next, we shall determine the net force acting on the mass. This can be obtained as follow:

Force applied (Fₐ) = 300 N

Force of friction (Fբ) = 30 N

Net force (Fₙ) =?

Fₙ = Fₐ – Fբ

Fₙ = 300 – 30

Fₙ = 270 N

Thus, the net force acting on the mass is 270 N.

Finally, we shall determine acceleration of the mass. This can be obtained as follow:

Mass (m) of object = 40 Kg

Net force (Fₙ) = 270 N

Acceleration (a) =?

Net force = mass × acceleration

Fₙ = m × a

270 = 40 × a

Divide both side by 40

a = 270 / 40

a = 6.75 m/s²

Therefore, the acceleration of the mass is 6.75 m/s²

Answer:

True

Explanation:

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 5000 × 15

We have the final answer as

Hope this helps you

Answer:

122444 Kg

Explanation:

Using the formula for gravitational force;

F= Gm1m2/r^2

Given that

G = constant of gravitation = 6.67 × 10−11 newton-metre2-kilogram−2.

m1 and m2 = the masses

r= distance of separation

Note that the question specifically mentioned that the masses are identical

so m1=m2 = m^2

F= Gm^2/r^2

Fr^2 = Gm^2

m^2 = Fr^2/G

m^2 = 1 × 1^2/6.67 × 10−11:

m =√ 6.67 × 10−11

m= 122444 Kg

Answer:

The correct answer is - the sound waves make vibration that travels through the string.

Explanation:

When an individual person talks into your paper cup telephone the person on the other end can feel the bottom of their cup vibrate. The sound waves create vibration go through the string that travels through the string to the end of the cup where vibrations can feel.

The sound waves are longitudinal waves that move or travel through different mediums like air, solid, or gas. The waves create vibration in the particles.

You could create a paper cup telephone but instead of using string, test out different materials and see if those materials will allow sound vibrations to travel through them

Answer:

1.429*10^-5 m

Explanation:

From the question, we are given that

Diameter of the cable, d = 3 cm = 0.03 m

Force on the cable, F = 2 kN

Young Modulus, Y = 2*10^11 Pa

Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²

The fractional length = Δl/l

Δl/l = F/AY

Δl/l = 2000 / 0.0007 * 2*10^11

Δl/l = 2000 / 1.4*10^8

Δl/l = 1.429*10^-5 m

Therefore, the fractional length is 1.429*10^-5 m long

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

          F = [tex]- \frac{m_{1} m_{2} }{r^2}[/tex]

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

rocket force -Earth

          F₁ = - \frac{m' M }{r^2}

rocket force - Moon

          F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

          2 F₁ = F₂

let's substitute

          2 [tex]2 \frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]

          (d-r) ² = [tex]\frac{m}{2M}[/tex] r²

           d² - 2rd + r² = \frac{m}{2M} r²

           r² (1 -\frac{m}{2M}) - 2rd + d² = 0

Let's solve this quadratic equation to find the distance r, let's call

           a = 1 - \frac{m}{2M}

           a = 1 - [tex]\frac{7.36 10^{22} }{2 \ 5398 10^{24}}[/tex] = 1 - 6.15 10⁻³

           a = 0.99385

            a r² - 2d r + d² = 0

           r =  [tex]\frac {2d \frac{+}{-} \sqrt{4d^2 - 4 a d^2}} {2a}[/tex]

           r = [2d ± 2d [tex]\sqrt{1-a}[/tex]] / 2a

           r = [tex]\frac{d}{a}[/tex]   (1 ± √ (1.65 10⁻³)) =  [tex]\frac{d}{a}[/tex] (1 ± 0.04)

           r₁ = \frac{d}{a} 1.04

           r₂ = \frac{d}{a} 0.96

let's calculate

           r₁ = [tex]\frac{3.84 10^8}{0.99385}[/tex] 1.04

           r₁ = 401.8 10⁸ m

          r₂ = \frac{3.84 10^8}{0.99385} 0.96

          r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m