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BARSIC [14]
3 years ago
6

An object is pulled with 20 net of force to the right against a friction force of 10 net.what is the net force.in what direction

is The friction force
Physics
1 answer:
Alenkasestr [34]3 years ago
6 0
Yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy
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Kinetic, thermal and electrical. There is more then one form of energy
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The equation for the chemical reaction shown is not balanced. What number should replace the question mark to balance this equat
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Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop using your fi
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3 years ago
An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the
vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>

<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>

6 0
2 years ago
A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are
Degger [83]

Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, \theta=30^{\circ}

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)

v = 0

Fd\ cos\theta=\dfrac{1}{2}m(u^2)      F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N

The force required to push to stop the car is 288.67 N

3 0
2 years ago
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