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PolarNik [594]
3 years ago
8

What is the speed and velocity of a car that travels adistance of 2000 m in 40 seconds north?​

Physics
1 answer:
Vanyuwa [196]3 years ago
7 0

Answer:

50m/s

Explanation:

Solution: Here,given;

distance covered (s) = 2000 meters

time taken (t) = 40 seconds

velocity (V) = ?

speed (V) = ?

Velocity(V) = Distance travelled (s)/Time taken (t)

or, V = 2000/40

or, V = 50m/s

V = 50m/s

hence, Velocity (V) = Speed (V)

therefore Speed (V) = 50 m/s

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A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire
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Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance =  20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

W_{fr} = -F\cdot d

Where, F = force

d = distance

Put the value into the formula

W_{fr}=-35.0\times20

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(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

W=fd\cos\theta

Put the value into the formula

W=35.0\times20\cos90

W=0\ J

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

W_{sh}=W_{net}-W_{fr}

Put the value into the formula

W_{sh}=0-(-700)

W_{sh}=700\ J

(D). We need to calculate the force the shopper exerts

Using formula of force

F_{sh}=\dfrac{W_{fr}}{d\cos\theta}

Put the value into the formula

F_{sh}=\dfrac{700}{20\cos25}

F_{sh}=38.61\ N

(E). We need to calculate the total work done on the cart

Using formula of work done

W_{cart}=W_{fr}+W_{sh}

Put the value into the formula

W_{cart}=700-(-700)

W_{cart}=0\ J

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

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