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zhannawk [14.2K]

4 years ago

9

A block of mass = 4.00 kg is supported by a spring scale with unit of measure of Newtons. This spring scale is attached to top o

f an elevator. The elevator accelerates upward with a = 3.00 m/sec2. What is the reading on the spring scale to 3 significant figures?

1 answer:

faust18 [17]4 years ago

5 0

Answer:

118 N

Explanation:

Given mass of the block, m = 4.00kg.

The acceleration of the elevator, a = 3.0 m/s^2.

As elevotar attaced with spring scale and accelerating upward

(block and elevator), so total force

$F_N-mg=ma$

Here, mg is the weight of the block downward direction.

or

$F_N=ma+mg=m(g+a)$

substitute the given value, we get

$F_N=4kg(9.8m/s^2+3m/s^2)$

= 117.6 N = 118 N.

Thus, the reading on the spring scale to 3 significant figures is 118 N.

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A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

GREYUIT [131]

Answer:

The flux through the surface of the cube is $2.314\ Nm^{2}/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0\times 10^{- 2}\ m$

Volume Charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$

Now,

To calculate the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$ (1)

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$ (2)

Volume of cube, $V = a^{3}$

Thus

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$

Thus from eqn (2), the total charge is given by:

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$

5 0

3 years ago

A pulse of light takes 3.00 ns to travel through air from an emitter to a detector. When a piece of transparent material with a

cupoosta [38]

Answer:

(a) $1.75\times10^8\text{ m/s}$

(b) 1.71

Explanation:

(a) The difference in the times of travel in the two case = $3.20 - 3.00 = 0.20\text{ ns} = 2.0\times10^{-9}\text{ s}$

This difference is the time in the transparent material. With a thickness of 35.0 cm, the speed in the material is

$v = \dfrac{35 cm}{2.0\times10^{-9}\text{ s}}=\dfrac{0.35 m}{2.0\times10^{-9}\text{ s}} = 1.75\times10^8\text{ m/s}$

(b) The refractive index of the material is the ratio of the velocity of light in vacuum to its velocity in the material. Using speed of light in vacuum as $c = 3.00\times10^8\text{ m/s}$ , the refractive index, $n$ , is

$n=\dfrac{c}{v} = \dfrac{ 3.00\times10^8\text{ m/s}}{1.75\times10^8\text{ m/s}}= 1.71$

4 0

3 years ago

PLEASE HELP ASAP!!! PLEASEEEEEEE

Andrew [12]

-- Top picture:

'a' and 'b' have the same magnitude, and both are positively charged.

-- Bottom picture:

The charge on the particle is negative.

8 0

3 years ago

On which moment of inertia of an object does depend on?

Likurg_2 [28]

Additional Information:

I couldn't get your question very clearly. In order to solve the question, I will define moment of inertia, state the formula and factors that the moment of inertia of a body depends and does not depend on.

Answer:

<u>Moment of inertia depends on;</u>

1. Mass of the body

2. Axis of rotation and

3. Distribution of the body

<u>Moment of inertia does not depend on;</u>

1. Angular velocity of the body.

Explanation:

The moment of inertia is defined as a quantity that determines the torque needed for a desired angular acceleration or a property of a body due to which it resists angular acceleration about a rotational axis.

Moment of Inertia, I = ∑mr²

Where,

I is the moment of Inertia

m is the mass

r is the distance from the axis of the rotation

The moment of inertia of a body depends on distribution of body, axis of rotation and mass of the body. However, the moment of Inertia of a body is not dependent on angular velocity of the body.

4 0

4 years ago

A 108 kg clock initially at rest on a horizontal floor requires a 639 N horizontal force to set it in motion. After the clock is

labwork [276]

Answer:

$\mu_s=0.60$

Explanation:

It is given that,

Mass of the clock, m = 108 kg

Force acting on it when it is in motion, $F=639\ N$

After the clock is in motion, a horizontal force of 521 N keeps it moving with a constant velocity, F' = 521 N

It is assumed to find the coefficient of between the clock and the floor. The force of friction is given by :

$F=\mu_smg$

$\mu_s=\dfrac{F}{mg}$

$\mu_s=\dfrac{639\ N}{108\ kg\times 9.8\ m/s^2}$

$\mu_s=0.60$

So, the coefficient of static friction between the clock and the floor is 0.6. Hence, this is the required solution.

5 0

3 years ago