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stiv31 [10]
3 years ago
12

A shopper at the Fries’ market place pushes a 13.0-kg shopping cart at a constant velocity for a distance of 58.7 m on a level s

urface. He pushes in a direction 24.50 below the horizontal. A 53.8 N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts on the cart? (b) What is the work done by the pushing force? (c) What is the work done by the frictional force? (d) What is the work done by the gravitational force?
Physics
1 answer:
Gnom [1K]3 years ago
4 0

We do a force summatory, then,

a) \sum F=0

Fcos(24.5)-F_f= 0

Fcos(24.5)-53.8= 0

F= \frac{53.8}{cos(24.5)}

F=59.12 N

(b) For the Work we need the equiation with the variable in X.

W= F (d cos(24.5))

W= 59.12 (58.7) cos 24.5

W=3158.1 J

(c) However the Frictional Force Work,

W= F_f d

W= -53.8(58.7)

W=-3158.1 J

(d) Now the work by gravity

W_g= F d (cos(90))=0

W_g=0

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Answer:

The maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

Explanation:

Let

m = Mass of Tarzan

M = Mass of Jane

Given

M = 2m

To calculate the maximum height Tarzan and Jane can swing, we make use of the potential energy at their initial and final position.

Reason being that;

At both the initial and final position, velocity is 0, so there's no kinetic energy.

And the potential energy remains the same (i.e constant) at any given point in the system.

Using P.E = mgh.

At initial position, PE1 = mgh

At final position, PE2 = (m + M)gH.

Where h and H represent the initial and final heights.

m + M is the new weight after Jane and Tarzan swing

Equating PE1 to PE2

mgh = (m + M)gH

By substituton (M = 2m)

mgh = (m + 2m)gH

mgh = 3mgH

Make H the subject of the formula

H = mgh/3mg

H = ⅓h

Hence, the maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

From the question, the new height looks to be about ½ that of Jane's original position; i.e. ½h

The calculated height is smaller than what the cartoon is showing;

We can conclude that the cartoon is wrong.

4 0
3 years ago
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14 electrone-14

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7 0
2 years ago
Which is more dense lead or Gold ?
WARRIOR [948]

Answer:

Gold is more dense.

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Answer:

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Explanation:

let:

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λ = (3.2×10^-4 cm)×sin(25.0°)/3

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