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Viktor [21]
1 year ago
6

A person is dragging a crate over a shag carpet with a push force of 10N. The shag rug creates a friction force of 5N. The mass

of the crate is 5 k
g. Calculate the acceleration of the crate?
(Just enter the numbers, no
Physics
1 answer:
vovangra [49]1 year ago
5 0

The acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

<h3>What is acceleration?</h3>

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity with time.

The  acceleration of the crate is calculated from the net force acting on the crate as shown below;

F(net) = applied force - frictional force

F(net) = 10 N - 5N

F(net) = 5 N

Apply Newton's second law of motion to calculate the acceleration of the crate.

F(net) = ma

where;

  • m is mass of the crate
  • a is acceleration of the crate

a = F(net) / m

a = 5/5

a = 1 m/s²

Thus, the acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

Learn more about acceleration here: brainly.com/question/14344386

#SPJ1

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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
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Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y =  \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} }  =2\ m/s

Also:

\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x =  u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s

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A roundabout is a type of playground equipment involving a large flat metal disk that is able to spin about its center axis. A r
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Answer:

116.1 kgm²/s

1.12718 rad/s

Decreases

Explanation:

m = Mass of girl = 43 kg

M = Mass of roundabout = 120 kg

v = Velocity of roundabout = 2.7 m/s

r = Radius of roundabout = 1 m = R

I = Moment of inertia

Her angular momentum

L_i=mvr\\\Rightarrow L_i=43\times 2.7\times 1\\\Rightarrow L_i=116.1\ kgm^2/s

Magnitude of angular momentum is 116.1 kgm²/s

Here the angular momentum is conserved

L_f=L_i\\\Rightarrow I\omega=L_i\\\Rightarrow (\frac{1}{2}MR^2+mr^2)\omega=116.1\\\Rightarrow \omega=\frac{116.1}{\frac{1}{2}\times 120\times 1^2+43\times 1^2}\\\Rightarrow \omega=1.12718\ rad/s

Angular speed of the roundabout is 1.12718 rad/s

Initial kinetic energy

K_i=\frac{1}{2}mv^2\\\Rightarrow K_i=\frac{1}{2}43\times 2.7^2\\\Rightarrow K_i=156.735\ J

Final kinetic energy

K_f=\frac{1}{2}I\omega^2\\\Rightarrow K_f=\frac{1}{2}\times (\frac{1}{2}\times 120\times 1^2+43\times 1^2)\times 1.12718^2\\\Rightarrow K_f=65.43253\ J

The overall kinetic energy decreases as can be seen. This loss is converted to heat.

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