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Murljashka [212]
3 years ago
14

Which of the following quantities are vectors?

Physics
1 answer:
Arte-miy333 [17]3 years ago
3 0

Answer:

electric fields

Explanation:

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the resultant force
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What happens when the thermal energy of a substance increases?
lawyer [7]

Answer: What happens when the thermal energy of a substance increases is that D. Both the motion of the particles in the substance and the temperature of the substance increase.

Explanation: When thermal energy increases, it is due to particles moving faster. It also results in higher temperatures or heat.

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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

3 0
3 years ago
Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou
spin [16.1K]

Answer:

(a) 28690 m/s (b) 2.46x10^{33}J

Explanation:

The orbital speed is define as:

v = \frac{2 \pi r}{T}   (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

v = \frac{d}{t}   (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

d = vt   (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

8min . \frac{60s}{1min} ⇒  480s

t = 480s

Replacing all those values in equation 3 it is gotten:

d = (3.0x10^{8}m/s)(480s)

d = 1.44x10^{11}m

Kepler’s third law is defined as:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

T = \sqrt{r^{3}}

T = \sqrt{(1.44x10^{11}m)^{3}}

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

T = \sqrt{1AU}

T = 1AU

That can be expressed in units of years.

T = 1AU . \frac{1year}{1AU}

T = 1year

But there are 31536000 seconds in one year:

T = 1year . \frac{31536000s}{1year}

T = 31536000s

Finally, equation  1 can be used:

v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}

v = 28690 m/s

<u>So Earth orbital speed around the Sun is 28690 m/s.</u>

<em>b) What is its kinetic energy?</em>

The kinetic energy is defined as:

E = \frac{1}{2}mv^{2}  (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

F = \frac{GMm}{r^{2}}

ma = \frac{GMm}{r^{2}}  (5)

M will be isolated in equation 5:

M = \frac{r^{2}a}{G}

Where r is the radius of the Earth (6.38x10^{6}m)

M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})

M = 5.98x10^{24} Kg

E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}

E = 2.46x10^{33}Kg.m^{2}/s^{2}

E = 2.46x10^{33}J

<u>Hence, the kinetic energy of Earth is 2.46x10^{33}J.</u>

8 0
3 years ago
An athlete runs with velocity 11 km/h for 5 minutes, 16 km/h for the next 4 minutes, and 18 km/h for other 6 minutes. Compute th
podryga [215]

Answer:

46.656km

Explanation:

it covers 46.656 km

7 0
2 years ago
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