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OleMash [197]
3 years ago
10

A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficien

ts of static and kinetic friction are 0.697 and 0.371, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? If the block does not move, give 0 m/s2 as the acceleration?
Physics
1 answer:
Liula [17]3 years ago
7 0

Answer:

Yes it will move and a= 4.19m/s^2

Explanation:

In order for the box to move it needs to overcome the maximum static friction force

Max Static Friction = μFn(normal force)

plug in givens

Max Static friction = 31.9226

Since 36.6>31.9226, the box will move

Mass= Wieght/g which is 45.8/9.8= 4.67kg

Fnet = Fapp-Fk

= 36.6-16.9918

=19.6082

=ma

Solve for a=4.19m/s^2

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When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequenc
Arlecino [84]

Answer:

m_2/m_1=9.745

The ratio m_2/m_1 of the masses is 9.745

Explanation:

Formula for frequency when mass m_1 is hung on the spring:

f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}

where:

k is the spring constant

Formula for frequency when mass m_2 is hung on the spring along with m_1:

f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}

where:

k is the spring constant.

In order to find ratio m_2/m_1, Divide the above equations:

\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}

On Solving the above equation:

\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745

The ratio m_2/m_1 of the masses is 9.745

8 0
4 years ago
Two exactly similar wire of steel and copper are stretched by equal force.if the total elongation is 10cm.find how much each wir
antiseptic1488 [7]

Answer:

The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.

Explanation:

Young's modulus is defined as:

E = stress / strain

E = (F / A) / (dL / L)

E = (F L) / (A dL)

Solving for dL:

dL = (F L) / (A E)

The wires have the same force, length, and cross-sectional area.  So:

dL₁ + dL₂ = (FL/A) (1/E₁ + 1/E₂)

Given that dL₁ + dL₂ = 0.10 m, E₁ = 20×10¹⁰ N/m², and E₂ = 12×10¹⁰ N/m²:

0.10 = (FL/A) (1/(20×10¹⁰) + 1/(12×10¹⁰))

FL/A = 0.75×10¹⁰ N/m

Solving for dL₁ and dL₂:

dL₁ = (FL/A) / E₁

dL₁ = (0.75×10¹⁰ N/m) / (20×10¹⁰ N/m²)

dL₁ = 0.0375 m

dL₂ = (FL/A) / E₂

dL₂ = (0.75×10¹⁰ N/m) / (12×10¹⁰ N/m²)

dL₂ = 0.0625 m

The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.

5 0
3 years ago
This is an essay don't make it to long but please help its science
lawyer [7]
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4 years ago
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An ideal refrigerator does 240 J of work to remove 610 J as heat from its cold compartment. (a) What is the refrigerator's coeff
Anastaziya [24]

Explanation:

It is given that,

Work done, W = 240 J

Heat removed, Q = 610 J

(a) The refrigerator's coefficient of performance is given by :

COF=\dfrac{Q}{W}

COF=\dfrac{610}{240}

COF = 2.54

(b) Let Q' is the heat per cycle is exhausted to the kitchen. It is given by :

Q' = Q + W

Q' = 610 + 240

Q' = 850 J

Hence, this is the required solution.

4 0
4 years ago
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<span>A_____ is the amount of heat needed to raise the temperature of 1 kilogram of water 1 degree celsius.


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