Answer:
M = 0.730*m
V = 0.663*v
Explanation:
Data Given:
Conservation of Momentum:
Energy Balance:
Substitute Eq 2 into Eq 1
Using Eq 1
A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient
of sliding friction between the sidewalk and the metal runners of the sled?
1 answer:
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
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Answer:
Part a)
Part c)
This mean the truck is decelerating and its speed is decreasing
Part e)
Part f)
Explanation:
Part a)
Speed of the truck is given as
as we know that
1 km = 1000 m
1 h = 3600 s
so we will have
Part c)
Average acceleration is given as
now we have
Part e)
as we know that it is uniform acceleration
so we can say
Part f)
distance traveled by the truck is given as
Answer:
Approximately at that distance from the center of the planet.
Option A) The low value of near the cloud top of Saturn is possible because of the low density of the planet.
Explanation:
The value of on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:
,
where
.
is the mass of the planet, and
is the mass of the object.
To find an equation for , divide the equation for gravity by the mass of the object:
.
In this case,
, and
.
Calculate based on these values:
.
Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.
Refer to the equation for :
.
The mass of the planet is in the numerator. If two planets are of the same size, would be greater at the surface of the more massive planet.
On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since is in the denominator of
, increasing the value of
while keeping
constant would reduce the value of
. That explains why the value of
near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately
.
As a side note, likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of
near the cloud tops of Saturn is approximately
.
observer is standing at distance d = 60 m south from the intersection
cyclist is travelling at speed v = 10 m/s
now after t = 8 s its displacement from intersection is given by
so the position of cyclist makes an angle with the observer
now the component of velocity of cyclist along the line joining its position with the observer is given as
here
so at this instant cyclist is moving away with speed 8 m/s
Find the given attachment for solution
