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Harman
1 month ago
13

A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1) . The board rests on a frictionless horizontal s

urface. The coefficient of static friction between the board and the box is ?s. The coefficient of kinetic friction between the board and the box is, as usual, less than ?s.
Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.

Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

Express your answer in terms of some or all of the variables ?s, m1, m2, g, and L. Do not include Ff in your answer.
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
3 0

Explanation:

The entire system will accelerate due to the applied force. The box will feel a force opposing friction, and once this force surpasses the friction, the box will start moving. Therefore,

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The force applied is expressed as

F = (m1 + m2)×a hence

F = μs×g×(m1+m2)

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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
serg [3582]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: The speed of a wave on a string under tension can be determined using the following:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

Distance a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5×10^{-6}  m.

When tension is doubled:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Distance in the same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With the increased tension, it moves \Delta x = 15.4×10^{-5} m

4 0
1 month ago
A 2.1 × 103 kg car starts from rest at the top of a driveway that is sloped at an angle of 20.0° with the horizontal. An average
Keith_Richards [3271]

Answer:

The driveway measures 4.98 m

Explanation:

We aim to find the length of the driveway, thus utilizing the following equations

W=ΔK.E    where W represents work and  ΔK.E   indicates the change in kinetic energy

Moreover,

K.E = \frac{ MV^2}{2}also

W = F.d  where F is the force and d denotes distance

Given that

= 4000 N indicating this frictional force

F_{f} m = 2100 Kg  

θ= 20.0°  

V=3.8 m/s representing the car's speed at the bottom of the driveway

W=Δ K.E

=  15162  J  

W = (1/2)(2100)(3.8)^2As the x component of gravity is

= mg sinФ

Fx_thus

= (2100)(9.8)sin(20.0°) results in

Fx_{} = 7038.77 N

And the Net force isFx_{}

=

-

F_{net}Fx_ {} = 7038.77 - 4000 = 3038.77 NF_{f}

So, the length of the driveway equals W / (

) = 15162/3038.77 = 4.98 m F_{net}

F_{net}

Thus, this is the length of the driveway.

3 0
11 days ago
A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction o
Maru [3345]

Answer:

The direction in which a vehicle accelerates aligns with its velocity direction. However, the force of acceleration works against the car's speed.

Explanation:

The car’s initial acceleration can be found using:

v = v₀ + a t

a = (v-v₀) t

which assumes the initial speed is zero (v₀ = 0 m/s).

a = v / t

a = 300 / t

The acceleration vector matches the direction of the vehicle's movement.

Upon hitting the wall, a force is exerted in the reverse direction to halt the car, thus this acceleration opposes the vehicle’s speed. However, the module should be much greater since the stopping distance is minimal.

5 0
1 month ago
A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite
inna [3103]

Answer:

a. Angle= 28.82°

b. Approved. Although he might feel cold, he should be able to cross.

Explanation:

Velocity Vector

Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).

a.

The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

v_{ey}=v_r

We can calculate the vertical component of the explorer's swimming speed as

v_{ey}=|v_e|cos\alpha

Thus

v_r=|v_e|cos\alpha

Finding the value of \alpha

\displaystyle cos\alpha=\frac{v_r}{|v_e|}

\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876

Then the angle is given by

\alpha=28.82^o

b.

The component of the explorer's velocity that goes horizontally is

v_{ex}=0.759sin28.82^o

v_{ex}=0.366\ m/s

This represents the actual velocity directed towards the campsite

Considering that

\displaystyle v=\frac{x}{t}

To find t

\displaystyle t=\frac{x}{v}

Calculating the duration for the explorer to cross the river

\displaystyle t=\frac{29.3}{0.366}

t=80\ sec

As this time is under the hypothermia threshold (300 seconds), the conclusion is

Approved. Although he will feel cold, he should manage to cross successfully.

3 0
1 month ago
A snapshot of three racing cars is shown in the diagram. All three cars start the race at the same time, at the same place, and
Maru [3345]

Answer:

The car that is the furthest from the finish line is: Car III (Choice C).

Explanation:

Here, we seek the car with the lowest overall average speed throughout the race. Thus, the one in last place inherently possesses the slowest average speed.

Since Car III is significantly behind Cars I and II, Choice A and B cannot be correct. Choice D is also not valid, as the positions of the cars are not the same. Lastly, Choice E is incorrect due to sufficient evidence demonstrating that Choice C has the lowest average speed.

8 0
2 months ago
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