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Zepler [3.9K]
3 years ago
8

A 2.0-N force acts horizontally on a 10-N block that is initially at rest on a horizontal surface. The coefficient of static fri

ction between the block and the surface is 0.50. Suppose that the block now moves across the surface with constant speed under the action of a horizontal 3.0-N force. Which one of the following statements concerning this situation is false? The net force on the block is zero newtons. The coefficient of kinetic friction between the block and the surface is 0.30. The frictional force on the block has magnitude 3.0 N. The direction of the total force that the surface exerts on the block is vertically upward. The block is not accelerated.
Physics
1 answer:
Nady [450]3 years ago
7 0

Answer:

None of the statement is false - all of them are true

Explanation:

Let's analyze each statement:

- The net force on the block is zero newtons. --> TRUE. In fact, the block is moving across the surface at constant speed: constant speed means zero acceleration, and according to Newton's second law,

F = ma

this also means a net force of zero newtons.

- The frictional force on the block has magnitude 3.0 N. --> TRUE. The horizontal acceleration of the block is zero, so the resultant of the horizontal forces must be zero:

F-F_f = 0

where F = 3.0 N is the horizontal push and F_f is the frictional force. From the equation, we find

F_f = F = 3.0 N

- The coefficient of kinetic friction between the block and the surface is 0.30. --> TRUE. The frictional force is 3.0 N, and its expression is

F_f = \mu_k (mg)

where \mu_k is the coefficient of kinetic friction and (mg)=10 N is the weight of the block. Solving for \mu_k, we find

\mu_k = \frac{F_f}{mg}=\frac{3 N}{10 N}=0.30

- The direction of the total force that the surface exerts on the block is vertically upward. --> TRUE. Since gravity is acting downward, and the block is not accelerating on the vertical direction neither, there must be an equal and opposite force acting upward on the block: and this force is the force exerted by the surface on the block.

- The block is not accelerated. --> TRUE: the block is moving at constant speed, so its acceleration is zero.

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Explanation:

Let i be the angle of incidence and r be the angle of refraction .

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sin 72.65 / sinr = 1.333

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r = 46⁰

From the figure

Tan r  = d / 4

Tan 46 = d /4

d = 4 x Tan 46

= 4 x 1.0355

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3 years ago
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3 years ago
A 49 N coconut is in a 15 m tree. What is te potential energy of the coconut?
eduard

Answer:

735 J

Explanation:

From the question given above, the following data were obtained:

Weight (W) = 49 N

Height (h) = 15 m

Potential energy =?

Potential energy is simply defined as the product of weight of the object and height to which the object is raised. Mathematically, it is expressed as:

Potential energy = weight × height

With the above formula, we can obtain the potential energy of the coconut as follow:

Weight (W) = 49 N

Height (h) = 15 m

Potential energy =?

Potential energy = weight × height

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Thus, the potential energy of the coconut is 735 J

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A car moves with an average speed of 75 kmh^-1 from town P to town Q in 2 hours. By using information, you may calculate the dis
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Answer:

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Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

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Explanation:

patron

Explanation:

 The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.

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7 0
1 year ago
Read 2 more answers
A spherical bowling ball with mass m = 3.4 kg and radius R = 0.113 m is thrown down the lane with an initial speed of v = 8.1 m/
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Answer:

1. α = 67.28 rad/s²

2. a = -3.04 m/s²

3. t = 0.76 s

4. x = 5.28 m

5. vf = 5.78 m/s

Explanation:

1. Let's use the torque definition: τ = Iα.

The inertial moment of a sphere is I = (2/5)*m*R²

And we know that the torque is the cross product between force and distance, so we would have τ = FxR=|F|*|R|*sin(90)=|F|*|R|=μ*mg*R

Using these two definitions, we have: (2/5)*m*R²*α = μ*mg*R

So the magnitude of the angular acceleration would be: α = (5/2R)*μ*g = 67.28 rad/s².

2. The force definition is F = m*a, when a is the linear acceleration.

F = -μ*mg.

Then -μ*mg = m*a. Solving the equation for a we have: a = -μ*g = -3.04 m/s².

3. To get the time when the ball star to rolling we need to use angular and linear velocity equation.

- ωf = ω0 + α*t ; we assume that initial angular velocity is 0.

- vf = v0 - a*t; v0 is the initial linear velocity

The relation to pure rolling is: v = ω*R. Rewriting this equation in terms of time  v0 - a*t = α*t*R, so t = v0/(α*R+a) = 0.76 s.

4. Using the distance equation: xf = x0 + v0*t - 0.5*a*t² = 5.28 m.

5. vf = v0 - a*t = 5.78 m/s.

Have a nice day!

8 0
3 years ago
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