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galina1969 [7]
3 years ago
10

A car travels 1008km in 14 hours. What is the average speed of the car?

Physics
1 answer:
Montano1993 [528]3 years ago
6 0

Answer:

72 km/hr. ...jshdhdhddjidididdiididudd

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Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas
Westkost [7]

Answer:

(I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

Explanation:

Given that,

Mass of black hole m= 1\times10^{8} M_{sun}

(I). We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=2.94\times10^{8}\ km

(II). Mass of block hole m= 6 M_{sun}

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=17.7\ km

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}

R_{g}=1.1\times10^{-7}\ km

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}

R_{g}=7.4\times10^{-29}\ km

Hence, (I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

8 0
3 years ago
How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R
VladimirAG [237]

Answer:

This relationship is explained by Ohm's law

Explanation:

Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as

V= IR

8 0
3 years ago
How to do it? Urgent ​
mixer [17]

a)

F= ma

a=v/t

F=5*(35/5)

F=35N

b)

a=F/m

a=(35-2)/5

a=33/5

a=6.6N

8 0
3 years ago
The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan
34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0
3 years ago
A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
3 years ago
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