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ollegr [7]
2 years ago
8

A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa

gon handle, which is inclined at 35° to the sidewalk as shown below. What is the magnitude of the fore of friction on the wagon?
Physics
1 answer:
Jlenok [28]2 years ago
5 0

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

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Go to the roller coaster simulation and click on the "launch" button. Pay attention to the pie chart as the roller coaster moves
Paladinen [302]

Answer:

The kinetic energy INCREASES as the roller coaster goes downhill.

Kinetic energy is greatest at POINT 2

Potential energy is greatest at POINT 1

Kinetic energy is decreasing while potential energy is increasing between points 3 AND 4

Which chart comes closest to the relationship between kinetic energy and potential energy at point 6 - CHART OF ANY POINT IN THE SAME HEIGHT AS OF 6

Explanation:

⇒As the potential energy increases , kinetic energy decreases.

⇒Potential energy here is gravitational potential energy.

⇒Thus, more we move away from the centre of the earth , more will be the gravitational potential energy or decrease in kinetic energy

3 0
2 years ago
Read 2 more answers
A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?
Yuri [45]
W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.
5 0
2 years ago
Read 2 more answers
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so t
Serggg [28]

a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

L=I\omega

where

I is the moment of inertia

\omega is the angular speed

Since the angular momentum must be conserved, we can write

L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2

where we have

I_1 = 2.25 kg m^2 is the initial moment of inertia

\omega_1 = 5.00 rad/s is the initial angular speed

I_2 = 2.25 kg m^2 is the final moment of inertia

\omega_2 is the final angular speed

Solving for \omega_2, we find

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular speed

Applying the formula, we have:

- Initial kinetic energy:

K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J

- Final kinetic energy:

K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J

7 0
2 years ago
Determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding
stellarik [79]
Let there be N number of wires.

Maximum tension a wire can withstand = 100 lb

so, Total tension N wires can withstand =  100 N

now, total tension in N wires = Maximum weight of bucket

100 N  = W

so, W = 100N

W is the weight of bucket and 100N is its maximum value.
8 0
2 years ago
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