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Alborosie
3 years ago
13

consider a perfectly inelastic head-on collision between a small car and a large truck traveling at the same speed. Which vehicl

e has a greater change in kinetic energy as a result of the collision
Physics
1 answer:
sergeinik [125]3 years ago
3 0

Answer: The truck has a greater kinetic energy

Explanation:

The initial velocities of both the truck and car are the same. Likewise, the final velocity is the same. The change in kinetic energy is dependent only on mass. The truck has a greater mass, therefore, the change in its kinetic energy is greater.

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p=F/A

or,P=d×V×G/A (m=d×V)

or,p= d× A×h×g/A (A and A are cut)

or,P=d×H×G

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How did mount st. Helen get its name
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All matter in the Universe consists of many substances called elements. <br><br> true or faluse
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The answer to this statement is true!
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Which of the following statements about the impact of biotechnology on the environment is not true?
Vladimir79 [104]

E)  Biotechnology use can lead to an increase in soil erosion and pollution runoff through decreased pesticide use.

Explanation:

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There is no relationship between biotechnology use and increase soil erosion and pollution run off.

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To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
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