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spin [16.1K]
3 years ago
14

What is the mass of a ball that is moving at 25 m/s and has 93.75 kg*m/s of momentum ?

Physics
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

m = 3.75 [kg]

Explanation:

We must remember that momentum is defined as the product of mass by Velocity, therefore it can be represented by means of the following equation.

P=m*v

where:

P = momentum = 93.75 [kg*m/s]

m = mass [kg]

v = velocity = 25 [m/s]

Now replacing, we can clear the mass:

P=m*v\\m=P/v\\m=93.75/25\\m=3.75 [kg]

You might be interested in
Setting another object on top of it is this increasing or decreasing
Nostrana [21]

Answer:

Increasing

Explanation:

It’s increasing because your are adding more weight

4 0
3 years ago
A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0
Xelga [282]

Answer:

v_m \approx -4.38\; \rm m \cdot s^{-1} (moving toward the incline.)

v_M \approx 4.02\; \rm m \cdot s^{-1} (moving away from the incline.)

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

If g = 9.81\; \rm m \cdot s^{-2}, the potential energy of the block of m = 2.20\; \rm kg would be m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this m = 2.20\; \rm kg\! block right before the collision would also be approximately 77.695\; \rm J.

Calculate the velocity of that m = 2.20\; \rm kg based on its kinetic energy:

\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right before the collision: approximately 18.489\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right after the collision: (m\cdot v_m + m \cdot v_M).

For momentum to conserve in this collision, v_m and v_M should ensure that m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

Kinetic energy of the two blocks right before the collision: approximately 77.695\; \rm J and 0\; \rm J. Sum of these two values: approximately 77.695\; \rm J\!.

Sum of the energy of each block right after the collision:

\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right).

Similarly, for kinetic energy to conserve in this collision, v_m and v_M should ensure that \displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J.

Combine to obtain two equations about v_m and v_M (given that m = 2.20\; \rm kg whereas M = 7.00\; \rm kg.)

\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right..

Solve for v_m and v_M (ignore the root where v_M = 0.)

\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right..

The collision flipped the sign of the velocity of the m = 2.20\; \rm kg block. In other words, this block is moving backwards towards the incline after the collision.

6 0
3 years ago
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of
PtichkaEL [24]

Answer:

Heat is flowing into the metal.

Explanation:

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

Mass (M) of iron = 150 g

Initial temperature (T₁) = 25.0°C

Final temperature (T₂) = 73.3°C

Direction of heat flow =?

Next, we shall determine the change in the temperature of iron. This can be obtained as follow:

Initial temperature (T₁) = 25.0 °C

Final temperature (T₂) = 73.3 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 73.3 – 25

ΔT = 48.3 °C

Next, we shall determine the heat transfered. This can be obtained as follow:

Mass (M) of iron = 150 g

Change in temperature (ΔT) = 48.3 °C

Specific heat capacity (C) of iron = 0.450 J/gºC

Heat (Q) transfered =?

Q = MCΔT

Q = 150 × 0.450 × 48.3

Q = 3260.25 J

Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.

7 0
3 years ago
What has more momentum a large truck stopped at a stop sign or a motorcycle moving down the road.
Alona [7]
<h2>I'm pretty sure it's</h2><h2>"<u>a motorcycle moving down the road.</u>"</h2><h3></h3><h3>The truck isn't moving so the truck cannot have momentum.</h3><h3>Momentum is basically "mass in motion."</h3><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>
4 0
3 years ago
A boy lifts a 30N dragon 2 meters above the ground. How much work did the boy do in the dragon? What is the work?
tekilochka [14]

Work =FS

Work =30 x 2

Work =60J

7 0
3 years ago
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