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olasank [31]
3 years ago
9

A lawn mower is pushed with a force of 79 N. If 11,099 J of work are done on mowing the lawn, what is the total distance the law

n mower was pushed? Round your answer to a whole number (no decimal places).
Physics
1 answer:
Stells [14]3 years ago
4 0

Answer:

141m

Explanation:

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Velocity vector and acceleration vector in a uniform circular motion are related as.
mr_godi [17]

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9 months ago
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Geothermal power can provide consistent electricity throughout the day and year - continuous baseload power and flexible power to support the needs of variable renewable energy resources, such as wind and solar. Sustainable Investment.

Explanation:

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2 years ago
Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration
eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: -2336 m/s^2 on the hardwood floor, -382 m/s^2 on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

v^2 - u^2 = 2ad

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

d = 1.8 mm = 0.0018 m

So we can find the acceleration by using again the equation

v'^2 - v^2 = 2ad

Solving for a,

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2

For the carpeted floor instead,

d=1.1 cm = 0.011 m

therefore the acceleration is

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

a=\frac{v'-v}{t}

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

a=-2336 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms

For the carpeted floor,

a=-382 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

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2 years ago
Calculate the electric potential energy in a capacitor that stores 4.0 10-10 C of charge at 250.0 V
Arada [10]
Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
 = 4 × 10^(-10)/2.5 × 10^2
 = 1.6 × 10^(-12) volt
7 0
3 years ago
Read 2 more answers
A ball rolls off the end of a horizontal table that is 4 meters off the ground. It is measured that the ball lands 3 meters away
ElenaW [278]

Answer:

The speed at which the ball rolled off the end of the table is 3.3 m/s

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.

The position vector of the ball can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

When the ball reaches the ground, its position will be:

r final = (3, -4)

Then:

3 = x0 + v0x · t

-4 = y0 + v0y · t + 1/2 · g · t²

Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:

3 m = v0x · t

-4 m = 1/2 · g · t²

We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:

-4 m = 1/2 · g · t²

-4 m = -1/2 · 9.8 m/s² · t²

8 m / 9.8 m/s² = t²

t = 0.9 s

Then:

3 m = v0x · 0.9s

3 m/ 0.9 s = v0x

v0x = 3.3 m/s

The speed at which the ball roll off the end of the table is 3.3 m/s

8 0
3 years ago
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