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Alla [95]
3 years ago
12

The amount of friction divided by the weight of an object forms a unit less number called the

Physics
1 answer:
Romashka [77]3 years ago
3 0

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

F=\mu N

N is normal force.

\mu = coefficient of friction

\mu=\dfrac{F}{N}

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Which of the following statements are true concerning the reflection of light?
natka813 [3]

Answer:

b. The reflection of light from a smooth surface is called specular reflection.

c. The reflection of light from a rough surface is called diffuse reflection.

Explanation:

a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.

This is wrong: Based on law of reflection "The angle of incidence is equal to the angle of reflection when light strikes any plane surface" examples plane mirrors, still waters, plane tables, etc

b. The reflection of light from a smooth surface is called specular reflection.

This is correct

c. The reflection of light from a rough surface is called diffuse reflection.

This is correct

d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.

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For specular reflection, the angle of incidence is less than the angle of reflection.

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4 0
4 years ago
When 108 g of water at a temperature of 22.5?
FrozenT [24]
<span>ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC. Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used q=mCΔT 11466.58 Joules = 65.1g x 4.18 J / g C x ΔT 11466.58/(65.1gx4.18)=ΔT ΔT=42.14oC So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).</span>
3 0
4 years ago
Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is
marshall27 [118]

Answer:

x=0.0049\ m= 4.9\ mm

d=0.01153\ m=11.53\ mm

Explanation:

Given:

  • mass of the object, m=0.75\ kg
  • elastic constant of the connected spring, k=150\ N.m^{-1}
  • coefficient of static friction between the object and the surface, \mu_s=0.1

(a)

Let x be the maximum distance of stretch without moving the mass.

<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>

f=k.x

\mu_s.N=k.x

where:

N = m.g = the normal reaction force acting on the body under steady state.

0.1\times (9.8\times 0.75)=150\times x

x=0.0049\ m= 4.9\ mm

(b)

Now, according to the question:

  • Amplitude of oscillation, A= 0.0098\ m
  • coefficient of kinetic friction between the object and the surface, \mu_k=0.085

Let d be the total distance the object travels before stopping.

<em>Now, the energy stored in the spring due to vibration of amplitude:</em>

U=\frac{1}{2} k.A^2

<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>

U=F_k.d

\frac{1}{2} k.A^2=\mu_k.N.d

0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d

d=0.01153\ m=11.53\ mm

<em>is the total distance does it travel before stopping.</em>

7 0
3 years ago
1. What charge is stored in a 180.0-μFcapacitor when 120.0 V is applied to it?
marin [14]

Answer:

So capacitance is charge divided by voltage and we can multiply both sides by V to solve for Q. So Q—the charge stored in the capacitor—is the capacitance multiplied by the voltage. So it's 180 times 10 to the minus 6 farads times 120 volts which is 0.0216 coulombs.

Explanation:

3 0
3 years ago
Maya uses a tuning fork of frequency 250 Hz to produce transverse wave on a stretched string. Her friend justin measure the dist
Viktor [21]

Answer:6.25m/s

Explanation:

V=F*lander

V=FL/2=250*0.05/2

7 0
3 years ago
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