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pantera1 [17]
3 years ago
10

A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

e μs = 0.80 and μk = 0.60. A horizontal string is attached to the block and a constant tension T is maintained in the string. What is the force of friction acting on the block for each of the following values of T?
a) T=15N

b) T=35N
Physics
1 answer:
Georgia [21]3 years ago
8 0

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

a)

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

b)

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

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Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2

Let the tensile modulus of Nickel E = 170 \times 10^9Pa.

The elongation of the rod can be calculated using the following formula:

\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m

6 0
3 years ago
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

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The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 360 V , which charges a capacitor
aksik [14]

Answer:

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Explanation:

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By definition we know that power is represented as

P = \frac{E}{t}

Where,

E= Energy

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Solving to find the Energy we have,

E = P*t

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7 0
3 years ago
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