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3 years ago

The speed of light is about 300,000 km/s (3×105km/s). If Jupiter is 0.72 light hours from the Sun, how far is this?

Physics

2 answers:

Daniel [21]3 years ago

7 0

Answer:

7.77 x 10⁸ Km

Explanation:

given,

Speed of light = 3 x 10⁵ Km/s

= 3 x 10⁸ m/s

1 light hour = 1.079 x 10⁹ Km

now,

0.72 light hour = 0.72 x 1.079 x 10⁹ Km

= 0.7769 x 10⁹ Km

= 7.77 x 10⁸ Km

The Jupiter is 7.77 x 10⁸ Km far from sun.

frozen [14]3 years ago

7 0

Answer:

Explanation:

speed of light, v = 3 x 10^8 m/s

time = 0.72 hour

distance = velocity x time

distance = 3 x 10^8 x 0.72 x 3600

distance = 7776 x 10^8 m = 7.78 x 10^8 km

Thus, the distance is 7.78 x 10^8 km.

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Light travels from the sun to the earth, a distance of 144,000,000,000 meters, in 480 seconds. Calculate the approximate velocit

mel-nik [20]

Answer:

The answer is C.) 300,000,000 m/s

Explanation:

Light travels at a constant rate of 300,000,000 m/s. this can be determined by dividing the distance (144,000,000,000 meters) by the time (480 seconds). that's pretty fast. in fact, nothing can travel any faster than this. ever.

5 0

2 years ago

An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi

son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

$\frac{1}{2}*m*v^{2} =e*V$

where :

m is the mass of electron

v is the velocity

V is the potential difference

$v=\sqrt{\frac{2*e*V}{m} }$ eq 1

Radius of electron moving in magnetic field is given by:

$R=\frac{m*v}{q*B}$ eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

$R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}$

$B=\sqrt{\frac{2*m*V}{e*R^{2} } }$

$B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3}) }{(1.60*10^{-19})*(0.170)^{2} } }$

B=9.1397*10^-4 Tesla

3 0

2 years ago

A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.

Nastasia [14]

<h3>Solution for the above question : -</h3>

Ohm's law states that :

$v = ir$

the terms used are :

$r = resistance$
$v = potential \: \: difference$
$i = current$

let's solve for electric current :

$2 = i \times 40$

$i = \dfrac{2}{40}$

$i = 0.05 \: A$

$i = 50 \: mA$

$\mathfrak{good\: \: luck \: \: for \: \: your \: \: assignment}$

8 0

2 years ago

HELP ME PLEASE ILL MAKE U BRANLEIST IF WRITE

katen-ka-za [31]

so the 1st on is the one on the left, middle is right and the 3rd one is the right one

3 0

3 years ago

Read 2 more answers

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

2 years ago