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zimovet [89]
3 years ago
13

A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds?

Physics
2 answers:
sveta [45]3 years ago
5 0


Initial velocity = 20m/s
Accelerations = 3m/s^2
Time taken = 5s


Using second equation of motion
ut+(1/2)at^2


Where, s is the distance covered by the body
u is initial velocity
a is acceleration and t is time taken by that body

= (20)(5)+(1/2)(3)(5)^2
=100 + 37.5
= 137.5


So, the total distance covered by the body in next 5s would be 137.5m
Arlecino [84]3 years ago
5 0

The Correct answer to this question for Penn Foster Students is: 137.5 m

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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
3 years ago
4748.36242611007 round to the nearest tenth
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Answer:4748.36

Explanation:

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Answer:

Explanation:

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We need to find the current, I, and since we know Q and t we can substitute these values into the given equation:

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If we substitute these values, we will get I as:

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