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MA_775_DIABLO [31]
3 years ago
8

A block whose weight is 45 N rests on a horizontal table. A horizontal force of 36 N is applied to the block. The coefficient of

kinetic and static friction are 0.65 and 0.42, respectively. Will the block move under the influence of the force, and if so, what will be the blocks acceleration?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:

1.5 m/s²

Explanation:

For the block to move, it must first overcome the static friction.

Fs = N μs

Fs = (45 N) (0.42)

Fs = 18.9 N

This is less than the 36 N applied, so the block will move.  Since the block is moving, kinetic friction takes over.  To find the block's acceleration, use Newton's second law:

∑F = ma

F − N μk = ma

36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a

6.75 N = 4.59 kg a

a = 1.47 m/s²

Rounded to two significant figures, the block's acceleration is 1.5 m/s².

Usually the coefficient of static friction is greater than the coefficient of kinetic friction.  You might want to double check the problem statement, just to be sure.

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hram777 [196]

To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T  +m_w L_v

Here,

m_{Al}= Mass of Aluminum

C_p= Specific Heat of Aluminum

C_w= Specific Heat of Water

m_w = Mass of water

L_v =Latent of Vaporization

Replacing,

\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)

Converting,

\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})

\Delta Q = 440.409kCal

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8 0
3 years ago
Please help with these physics problems 1. A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to
Tatiana [17]

Answer:

1.) h = 164.8 m

2.) U = 49.1 m/s

3.) t = 1.43 seconds

Explanation:

1.) A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to the ground. The height of the building is...? 

Since the soccer ball is dropped from the building, the initial velocity U will be equal to zero

Using second equation of motion

h = Ut + 1/2gt^2

Substitutes the time into the formula

h = 1/2 × 9.8 × 5.8^2

h = 164.8 m

2. The Falcon 9 launches to a height of 123 meters. What is its vertical initial velocity?

At maximum height final velocity = 0

Using the third law of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8 × 123

U^2 = 2410.8

U = 49.1 m/s

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Since the apple fall from rest, the initial velocity U will be equal to zero

Using the second equation of motion,

h = Ut + 1/2gt^2

substitute all the parameters into the formula

10 = 1/2 × 9.8 × t^2

10 = 4.9t^2

t^2 = 10/4.9

t^2 = 2.04

t = 1.43 seconds

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3 years ago
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Answer:

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Answer:

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3 years ago
A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
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Simplifying and solving, we find the value of the acceleration:
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Now we can use the following relationship to find the distance covered by the skier before stopping, S:
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where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
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