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harkovskaia [24]
3 years ago
6

If a car accelerates at 7.5 mph/s, what would be its final speed after 5.0 seconds if its

Physics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

87.5mph

Explanation:

Given parameters:

Acceleration  = 7.5mph/s

Starting speed  = 50mph

Time  = 5s

Unknown:

Final speed  = ?

Solution:

Acceleration is defined as the rate of change of velocity with time:

   Acceleration  = \frac{Final speed  - Starting speed }{time }

Insert the parameters and solve;

 7.5  = \frac{Final speed  - 50 }{5 }

  Final speed - 50  = 7.5 x 5

  Final speed  = 7.5 x 5 + 50  = 87.5mph

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A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How
stepan [7]

Answer:

6 \times 10^4 \; \rm J.

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2, where

  • m is the mass of the car, and
  • v is the speed of the car.

The 2 \times 10^3\; \rm kg car would have an initial KE of:

\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J.

The 3 \times 10^3\; \rm kg car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum p of an object is equal to its mass m times its velocity v. In other words, p = m\cdot v.

Let the mass of the two cars be denoted as m_1 and m_2, and their initial speeds v_1 and v_2. Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as v_3.

Initial momentum of the two-car system:

\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}.

After the collision, both car would have a velocity of v_3 (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3.

Since momentum is conserved during the collision, the momentum of the system after the collision would also be 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}. That is: (m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}.

Solve for v_3:

\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}.

Hence, the total kinetic energy after the collision would be:

\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}.

The amount of kinetic energy lost during the collision would be:

\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}.

5 0
3 years ago
As a rubber band and moves forward, which of the following is true
Yuki888 [10]

Answer:

It can go back to it's original shape

Explanation:

7 0
3 years ago
Some help me on this please
satela [25.4K]

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.

8 0
3 years ago
Would a wire carrying more electrons transfer more energy than a wire carrying fewer electrons? Explain why or why not.​
Burka [1]

Answer:

no

Explanation:

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5 0
3 years ago
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Ansthethethethethe

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2 years ago
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