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mojhsa [17]
3 years ago
11

A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the

ground if the ball is thrown forward
Physics
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer:

The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s

Explanation:

Given;

speed of the ball thrown inside boxcar, V_B = 23.2 m/s

speed of the boxcar moving along the tracks, V_T = 34.9 m/s

Determine the speed of the ball relative to the ground if the ball is thrown forward.

If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.

V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s

Therefore,  the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.

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The temperature of air changes from 0 to 10°C while its velocity changes from zero to a final velocity, and its elevation change
aliya0001 [1]

Answer:

Final velocity = 119.83 m/s

Final elevation = 731.9 m

Explanation:

We are told that temperature of air changes from 0 to 10°C

Thus;

Change in temperature; ΔT = 10 - 0 = 10°C

Also, its velocity changes from zero to a final velocity. Thus;

v1 = 0 m/s

v2 is unknown

Also, its elevation changes from zero to a final elevation.

So, z1 = 0 and z2 is unknown

Now, we want to find v2 and z2 when the internal, kinetic and potential energy are equal.

Thus Equating the formula for both kinetic and internal energy gives;

½m(v2² - v1²) = mc_v•ΔT

m will cancel out and v1 is zero to give;

v2² = 2c_v•ΔT

v2 = √(2c_v•ΔT)

Where c_v is specific heat of constant volume of air with a constant value of 718 J/Kg.K

Thus;

v2 = √(2 × 718 × 10)

v2 = √14360

v2 = 119.83 m/s

To find z2, we will equate potential energy formula to that of the internal energy.

Thus;

mg(z2 - z1) = mc_v•ΔT

m will cancel out and since z1 is zero, then we have;

z2 = (c_v•ΔT)/g

z2 = 718 × 10/9.81

z2 = 731.9 m

4 0
3 years ago
A dog has a mass of 20 kg. If the dog is pushed across the ice with a force of 40 N, what is its acceleration?
olasank [31]

Answer:

The acceleration is 2 m/s2.

Explanation:

We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:

F = m x a  ---> a= F/m

a = 40 N/20 kg                   <em>  1N= 1 kg x m/s2</em>

a= 40 kgx m/s2/ 20 kg

<em>a= 2 m/s2</em>

7 0
3 years ago
Do rolling down a grassy hill has kinetic energy or potential energy
Ann [662]

If you're moving, then you have kinetic energy.

If you're not at the bottom yet, then you still have
some potential energy left.

4 0
3 years ago
Read 2 more answers
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
a man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s
alex41 [277]

Answer:

h = 90.10 m

Explanation:

Given that,

A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m

The initial speed of the stone, u = 10 m/s

The path followed by the projectile is given by :

h(t)=-4.9t^2+10t+85 ....(1)

For maximum height,

Put dh/dt = 0

So,

\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s

Put the value of t in equation (1).

h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m

So, the maximum height of the stone is equal to 90.10 m.

5 0
3 years ago
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