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Sati [7]
3 years ago
9

8) Calculate the kinetic energy of a truck that has a mass of 6100 kg and is moving at 55 m/s.

Physics
1 answer:
Inessa [10]3 years ago
7 0

Answer:

<h2>9,226,250 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

v is the velocity

m is the mass

From the question we have

k =  \frac{1}{2}  \times 6100 \times  {55}^{2}  \\  = 3050 \times 3025 \\  = 9226250

We have the final answer as

<h3>9,226,250 J</h3>

Hope this helps you

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A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
emmainna [20.7K]

Complete Question

A person throws a pumpkin at a horizontal speed of   4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The  pumpkin's vertical displacement  is  H = 27 .67 \ m

The  pumpkin's vertical velocity when it hits the ground is  v_v__{f}} = 23.298 \  m/s

Explanation:

From the question we are told that

   The  horizontal speed is  v_h  =  4 m/s

    The horizontal distance traveled is  d =  9.5 \ m

The horizontal distance traveled is mathematically represented as

           S =  v_h * t

Where t is the time taken

substituting values

          9.5 =  4 * t

   =>     t =  \frac{9.5}{4}

            t = 2.38 \ sec

Now the vertical displacement is mathematically represented as

        H  =  v_v t  +  \frac{1}{2} a_v t^2

now the vertical velocity before the throw is  zero

    So

          H =  0 +  \frac{1}{2} (9.8) * (2.375)^2

          H = 27 .67 \ m

Now the final vertical velocity  is mathematically represented as

          v_v__{f}} =  v_v + at

  substituting values

             v_v__{f}} =  0 + (9.8)* (2.375)

            v_v__{f}} = 23.298 \  m/s

7 0
3 years ago
The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick
Travka [436]

Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.  

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions.  If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>

Therefore,

a. vf = v2 is wrong because vf is greater than v2

b. vf is less than v2 is wrong because vf is greater than v2

c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

4 0
3 years ago
Acceleration increases over time once a force is applied to the object. Determine the acceleration at 3.5 sec? A) 3 m/s2 B) 6 m/
Novay_Z [31]

Answer: Acceleration does not increase over time once a force is applied to the object. It depends on the force and the object's mass. If those don't change, then the acceleration is constant.

Explanation:. F = M • a

We don't know either of those numbers, so we can't answer the question.

4 0
2 years ago
Read 2 more answers
A solid weighs 200N in air, 150N in water and 170N in a liquid. Find relative density of solid, relative density of liquid and d
fomenos

Answer:

\rho_{s} = 4

\rho_{l} = 0.6

\rho{liq} = 600 kg/m^{3}

Given:

Weight of solid in air, w_{sa} = 200 N

Weight of solid in water, w_{sw} = 150 N

Weight of solid in liquid, w_{sl} = 170 N

Solution:

Calculation of:

1. Relative density of solid, \rho_{s}

\rho_{s} = \frac{w_{sa}}{w_{sa} - w_{sw}}

\rho_{s} = \frac{200}{200 - 150} = 4

2. Relative density of liquid, \rho_{l}

\rho_{l} = \frac{w_{sa} - w_{sl}}{w_{sa} - w_{sw}}

\rho_{l} = \frac{200 - 170}{200 - 150} = 0.6

3. Density of liquid in S.I units:

Also, we know:

\rho{l} = \frac{\rho_{liq}}{\rho_{w}}

where

= {\rho_{liq}} = density of liquid

= {\rho_{w}} = 1000 kg/m^{3} = density of water

Now, from the above formula:

0.6 = \frac{\rho_{liq}}{1000}

\rho{liq} = 600 kg/m^{3}

3 0
2 years ago
What will most likely happen with bar magnets aligned NS NS
likoan [24]

Answer:

repel

Explanation:

8 0
3 years ago
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