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3 years ago

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Relativistic velocity is of the order of _____ of velocity of light A- 1/15th of the velocity of light B-1/20th of the velocity

of light C-1/10th of the velocity of light

1 answer:

ratelena [41]3 years ago

4 0

Answer:

Relativistic velocity is of the order of 1/10th of the velocity of light

Explanation:

We define relativistic speed (or velocity) as a speed that is a significant fraction of the speed of light: c = 3*10^8 m/s

Such that for these speeds, the special relativity theory starts to apply (the relativity effects starts to apply).

Usually, we define relativistic speeds as those that are of the order (or larger) of c/10, which is one-tenth of the speed of light.

Then the correct option is C:

Relativistic velocity is of the order of 1/10th of the velocity of light

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One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

3 years ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

3 years ago

A 200g piece of iron is heated at 100C. It is than dropped into water to bring its temperature down to 22C. What is the amount o

gizmo_the_mogwai [7]

Answer:

6926.4J

Explanation:

Given parameters:

Mass of iron = 200g

Initial temperature = 100°C

Final temperature = 22°C

Unknown:

Amount of heat transferred to the water = ?

Solution:

The quantity of heat transferred to the water is a function of mass and temperature of the iron;

H = m c Ф

m is the mass of the iron

Ф is the change in temperature

C is the specific heat capacity of iron = 0.444 J/g°C

Now;

insert the parameters and solve;

H = 200 x 0.444 x (100-22)

H = 6926.4J

8 0

3 years ago

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i

timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

$I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2$

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

$I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2$

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

$I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2$

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

$I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}$

$\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}$

$\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s$

4)Kinetic energy before:

$E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J$

Kinetic energy after:

$E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J$

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) $a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2$

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0

3 years ago

Elevator is accelerating upward 3.5 M/S2 and has a mass of 300 KG. The force of gravity is 2940 N. What is the tension force pul

forsale [732]

Answer:

T = 3990 N

Explanation:

The free body diagram for the elevator consists of a tension force pointing up, and its weight pointing down. So the elevator's net force is:

F = T - 2940N

ad at the same time, using Newton's second law, we have that this net force should equal the elevator's mass (300 kg) times its acceleration (a):

T - 2940N = 300kg (3.5m/s^2)

then

T = 2940 N + 1050 N

T = 3990 N

8 0

3 years ago