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2 years ago

a 1900 kg car slows down from a speed of 25m/s to a speed of 15m/s in 6.25s. how much work was done on the car

Physics

1 answer:

ra1l [238]2 years ago

7 0

Answer:

$-3.8\cdot 10^5 J$

Explanation:

According to the work-energy theorem, the work done on the car is equal to its variation of kinetic energy:

$W=\Delta K=K_f -K_i = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$

where

m = 1900 kg is the mass of the car

vf = 15 m/s is the final speed

vi = 25 m/s is the initial speed

Substituting the data, we find

$W=\frac{1}{2}(1900 kg)(15 m/s)^2-\frac{1}{2}(1900 kg)(25 m/s)^2 =-3.8\cdot 10^5 J$

and the work is negative because it is done against the motion of the car (in fact, the car slows down, so it loses kinetic energy).

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Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

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$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

$2v_o^{2}=v_1^{2} + v_2^{2}$

Substituting 5 m/s for $v_o$ then

$2*(5^{2})= v_1^{2} + v_2^{2}$

$50= v_1^{2} + v_2^{2}$

Also, it’s known that $v_1+v_2=5$ hence $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

$50=25+v_2^{2}-10v_2+v_2^{2}$

$2v_2^{2}-10v_2-25=0$

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Substituting, $v_1=-1.83 m/s$

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

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2 years ago

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and

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Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

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Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

$\Delta p=m(v-u)\\\\\Delta p=mv$

The change in momentum is equal to the impulse delivered. So,

$J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s$

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2 years ago

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

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Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

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So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

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