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2 years ago

13

In its own reference frame, a box has the shape of a cube 1.5 m on a side. This box is loaded onto the flat floor of a spaceship

and the spaceship then flies past us with a horizontal speed of 0.90 c.

1 answer:

Harlamova29_29 [7]2 years ago

5 0

Volume of the box observed at the given speed is 7.89 m³

Given:

Length of side of box = 1.5 m

horizontal speed = 0.90 c

To Find:

Volume of box

Solution: The amount of space occupied by a three-dimensional figure as measured in cubic units (as inches, quarts, or centimeters)

The edge length of the cube according to the stationary observer is calculated as

a' = a x √ 1 - v^2/c^2

a' = 1.5 x √ 1 - (0.9c)^2/c^2

a' = 1.5 x 0.19

a' = 0.285 m

The volume of the cube is calculated as follows;

V = A x a'

V = (a)^2 x a'

V = (1.5)^2 x 0.285

V = 7.89 m³

Thus, the volume of the box observed at the given speed is 7.89 m³

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Which type of friction force acts when there is no motion?

rosijanka [135]

Its called static friction.

3 0

4 years ago

Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm.

jasenka [17]

Answer:

$P_{1}-P_{2}=12675Pa$

Explanation:

From the equation of continuity we know that:

$v_{1}A_{1}=v_{2}A_{2}\\Given \\r_{1}=2.0cm\\r_{2}=1.0cm\\v_{1}=1.3m/s\\Density of water p=1000kg/m^{3}\\Now\\A_{1}=\pi r_{1}^{2} \\A_{2}=\pi r_{2}^{2}\\ Therefore\\v_{2}=v_{1}\frac{A_{1}}{A_{2}}\\v_{2}=v_{1}\frac{\pi r_{1}^{2} }{\pi r_{2}^{2}}\\v_{2}=v_{1}\frac{r_{1}^{2} }{r_{2}^{2}}\\v_{2}=1.3*\frac{2.0^{2} }{1.0^{2} } \\v_{2}=5.2m/s\\$

From Bernoulli equation we know that:

$P_{1}+1/2pv_{1}^{2}+pgy_{1}=P_{2}+1/2pv_{2}^{2}+pgy_{2}\\$

Now assuming $y_{1}=y_{2}$

$P_{1}+(1/2)pv_{1}^{2}=P_{2}+(1/2)pv_{2}^{2}\\ P_{1}-P_{2}=1/2pv_{2}^{2} -1/2pv_{1}^{2}\\ P_{1}-P_{2}=1/2p(v_{2}^{2}-v_{1}^{2} )\\ P_{1}-P_{2}=1/2*1000(5.2^{2}-1.3^{2} )\\ P_{1}-P_{2}=12675Pa$

5 0

3 years ago

A current of 17.0 mA is maintained in a single circular loop of 2.00 \mathrm{~m} circumference. A magnetic field of 0.800T is di

poizon [28]

The magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$ .

We have current carrying single circular loop placed in a magnetic field

parallel to the plane of the loop.

We have to determine the Magnetic moment of the loop.

<h3>What is the formula to calculate the magnetic moment of loop?</h3>

The formula to calculate the magnetic moment of the loop is -

M = NIA

where -

N - Number of turns

I - Current in the loop

A - Area of the loop

According to the question, we have -

I = 17mA = 0.017A

Circumference (C) = 2 meters

B = 0.8 T

Now -

C = 2

2 $\pi$ r = 2

$\pi$ r = 1

r = $\frac{1}{\pi }$

r = 0.32

Using the formula -

M = NIA

M = $\pi$ NI $r^{2}$

M = 3.14 x 1 x 0.017 x 0.32 x 0.32

M = 3.14 x 4 x 0.017

M = 5.5 x $10^{-3}$ A $m^{2}$

Hence, the magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$

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3 0

1 year ago

When a 40-n force, parallel to the incline and directed up the incline, is applied to a crate on a frictionless incline that is

Genrish500 [490]

<span>F= ma force is directed on the working surface so 40 = m x 2 hence m must be 20 units needed are probably kg</span>

8 0

3 years ago

An AC adapter for a telephone-answering unit uses a transformer to reduce the line voltage of 120 V (rms) to a voltage of 9.0 V.

Illusion [34]

Answer:

1. The number of turns on the secondary output is 18

2. The root mean square power delivered to the transformer is 48 Watts.

Explanation:

A transformer is an electronic device that can be used for increasing or decreasing the value of a given voltage. It consists of primary coils and secondary coil of a definte number of turns. When voltage flows in the primary coil, it induces voltage in the secondary coil. The two types are: step-up and step down transformers.

1. For a given transformer,

$\frac{V_{s} }{V_{p} }$ = $\frac{N_{s} }{N_{p} }$

where $V_{s}$ is the value of the induced voltage in the secondary coil, $V_{p}$ is he voltage in the primary coil, and $N_{s}$ is the number of turns of the secondary coil, $N_{p}$ is the number of turns in the primary coil.

From the question,

$V_{s}$ = 9.0 V, $V_{p}$ = 120 V, $N_{p}$ = 240, $N_{s}$ = ?

So that,

$N_{s}$ = $\frac{V_{s}*N_{p} }{V_{p} }$

= $\frac{9 * 240}{120}$

= 18

The number of turns on the secondary output is 18.

2. $Power_{rms}$ = $I_{rms}$ × $V_{rms}$

= 0.4 × 120

= 48 W

The rms power delivered to the transformer is 48 Watts.

8 0

3 years ago