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2 years ago

what would happen to the temperature on mercury during the day if a sunspot was directly in line with it's surface?

Physics

1 answer:

QveST [7]2 years ago

3 0

Answer:

mind-bogglingly hot

Explanation:

If a sunspot were pointed right at Mercury in this way, the planet would become even more mind-bogglingly hot. Sunspots look dark in our visual spectrum, so we might suspect that they are colder (or at least less hot) than the rest of the Sun; but in fact, they are hotter.

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A car with mass 1600 kg drives around a flat circular track of radius 28.0 m. The coefficient of friction between the car tires

egoroff_w [7]

Answer:

v=15.24 m/s

Explanation:

Given that

Mass ,m= 1600 kg

radius ,r= 28 m

Coefficient of friction ,μ = 0.83

The radial force on the car when it takes turn

$F=\dfrac{mv^2}{r}$

The friction force on the car

Fr= μ m g

The condition for motion without losing traction

F= Fr

$\dfrac{mv^2}{r}=\mu\ m g$

v²=μ r g

$v=\sqrt{\mu\ r g}$

Now by putting the values

$v=\sqrt{0.83\times 28\times 10}\ m/s$ ( take g=10m/s²)

v=15.24 m/s

The speed of the car will be 15.24 m/s

5 0

3 years ago

How might scientists study something that they cannot observe directly?

yarga [219]

Answer:

The sun

Explanation:

if they get too close even the heat that radiates of of it would burn you or give sunburn bad and the sun is a shining ball of energy it is way to hot for a human to get close

8 0

3 years ago

you paddle a canoe with the force of 200N. you and the canoe have a combined mass of 97KG . what is the acceleration of the cano

erastovalidia [21]

Use Newton's (second) Law:
$F=ma$
with your data:
$200=97a$
$a=2m/s^{2}$

6 0

3 years ago

A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,00

Savatey [412]

Answer:

The answer is "Option B".

Explanation:

$r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\$

$=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\$

7 0

3 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

azamat

Answer:

a. Speed = 1.6 m/s

b. Amplitude = 0.3 m

c. Speed = 1.6 m/s

Amplitude = 0.15 m

Explanation:

The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:

frequency = $\frac{1}{2(2\ s)}$ = 0.25 Hz

The wavelength of the wave is the distance between consecutive crests of wave. Therefore,

Wavelength = 6.4 m

Now, the speed of the wave is given as:

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

Speed = 1.6 m/s

Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:

Amplitude = (0.5)(0.6 m)

Amplitude = 0.3 m

frequency = $\frac{1}{2(2\ s)}$ = 0.25 Hz

Speed = (Frequency)(Wavelength)

Speed = (0.25 Hz)(6.4 m)

Speed = 1.6 m/s

Amplitude = (0.5)(0.3 m)

Amplitude = 0.15 m

8 0

3 years ago