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3 years ago

A father uses a rope to pull a child on a sled at a constant speed. of the ones listed which forces are present?

Physics

1 answer:

Amanda [17]3 years ago

4 0

All the forces may be present.

The child on the sled has a weight- which is due to the force of gravity on the child and the sled.

The sled and the child exert a force on the ground equal to the combined weight of the sled and the child. The ground exerts a normal force on the sled.

The force used by the father to pull the sled is the applied force.

The sled slides on the ground and as a result, the force of friction exists between the ground and the sled, directed opposite to the direction of the motion of the sled.

The father pulls the sled using a rope. As a result, the rope is under Tension.

As the sled moves it also experiences a force of air resistance, which is dependent on the sled's speed.

However, since the father pulls the sled along with constant speed, the sum of all the forces acting on the sled is zero.

Since there is no movement in the upward or downward directions, the weight of the child and the sled is equal to the normal force acted upon the sled by the ground.

The force applied by the father on the rope is equal to the tension in the rope.

Friction and air resistance act opposite to the direction of motion of the sled. for the sled to move at constant speed, the tension in the rope must be equal to the sum of the forces due to friction and air resistance.

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Calculate the magnitude of the total impulse applied to the car to bring it to rest

n200080 [17]

Answer:

Total impulse = $mv$ = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

$|J|=|p_f-p_i|$

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

$p_i=mv$

The final momentum of the car is given as:

$p_f=m(0)=0$

Therefore, the impulse is given as:

$|J|=|p_f-p_i|=|0-mv|=|-mv|=mv$

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

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3 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

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