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Neko [114]
3 years ago
10

An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0

% of the speed of light. Its nuclear-powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of 15.9 years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth?
yr
(b) How far is the probe from Earth when its batteries fail, as measured by mission control?
ly
(c) How far is the probe from Earth, as measured by its built-in trip odometer; when its batteries fail?
ly
Physics
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

22.26 years

, 15.585 light years  , 11.13 light years

Explanation:

a)

t' = t/(\sqrt{1-(v/(c*v)/c)}

= 15.9/\sqrt{(1-0.7*0.7)}

= 22.26 years

b)

0.7*c*22.26 years

=15.585 light years  

c)

0.7*c*15.9

=11.13 light years

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What are the different types of Energy transformations and Energy Transfers?
Harman [31]

Explanation:

Here are some of the ways that energy can change (transform) from one type to another:

The Sun transforms nuclear energy into heat and light energy.

Our bodies convert chemical energy in our food into mechanical energy for us to move.

An electric fan transforms electrical energy into kinetic energy.

7 0
3 years ago
The leading cause of air pollution is?
stellarik [79]
Combustion Engine Exhaust----Cars, trucks, jet airplanes and other combustion engine vehicles cause air pollution. The exhaust from these contains carbon monoxide, nitrous oxide and gaseous oxide. This type of air pollution creates smog
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3 years ago
A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet
puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

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3 0
3 years ago
PLEASE HELP ME 45 POINTS
sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a (1)

Mass B

\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a (2)

Where:

T - Tension force in the cord, measured in newtons.

m_{A}, m_{B} - Masses of blocks A and B, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

a - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g

(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g

a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g

If we know that m_{A} = 5\,kg, m_{B} = 2\,kg and g = 9.807\,\frac{m}{s^{2}}, then the acceleration of the masses is:

a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)

a = 4.203\,\frac{m}{s^{2}}

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

T = m_{B}\cdot (a+g)

If we know that m_{B} = 2\,kg, g = 9.807\,\frac{m}{s^{2}} and a = 4.203\,\frac{m}{s^{2}}, then the tension force in the cord:

T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}  \right)

T = 28.02\,N

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

\Delta y  = \frac{1}{2}\cdot a\cdot t^{2} (3)

Where:

a - Net acceleration, measured in meters per square second.

t - Time, measured in seconds.

\Delta y - Covered distance, measured in meters.

If we know that a = 4.203\,\frac{m}{s^{2}} and \Delta y = 1.5\,m, then the time taken by the system is:

t = \sqrt{\frac{2\cdot \Delta y}{a} }

t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }

t \approx 0.845\,s

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

v = a\cdot t (4)

Where v is the final speed of the system, measured in meters per second.

If we know that a = 4.203\,\frac{m}{s^{2}} and t \approx 0.845\,s, then the final speed of the system is:

v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)

v = 3.551\,\frac{m}{s}

The final speed of the system is 3.551 meters per second.

8 0
2 years ago
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