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Citrus2011 [14]

3 years ago

11

Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much

work done on the weights? show full working out

1 answer:

shutvik [7]3 years ago

6 0

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

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A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l

MrRissso [65]

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

$F = -kx$

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

$F=mg$

Inserting this into Hooke's Law

$mg=-kx$

computing it for x,

$-x=mg/k$

This is the model which will tell the length of the spring against change in the mass located in the pan.

3 0

3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

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Which of the following is not a benefit of increased energy efficiency?

Rzqust [24]

From what I can see it's D, I did this by simply examining the other answers and seeing that they are beneficial, so, from that information, this one must not be.

7 0

3 years ago

A horse pulls a plow with a 242 N force for

prohojiy [21]

Answer: The answer is D

Explanation: i had the same question and i just guessed and got it first try

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vovikov84 [41]

Answer:

Explanation:

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3 years ago