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3 years ago

Weight of a man is less at coal mine.why?

1 answer:

7 0

If he is underground then there is less planetary mass below him and therefore less gravity. weight is just gravity's affect on you. his mass won't change.

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4. An object is moving with an initial velocity of 9 m/s. It accelerates at a rate of 1.5

Mazyrski [523]

Answer:

11.87 ms⁻¹

Explanation:

You can use the kinematic equation

v² = u² + 2as

Where v = final velocity

u = initial velocity

a = acceleration

s = displacement

v² = 9² + 2×1.5×20

So you get v = 11.87 ms⁻¹

7 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

What is the potential Energy?

Zinaida [17]

Answer:

potential energy is the energy of an object that is not moving

Explanation:

example a skateboarder stands at the top of the ramp he has no kinetic energy but alot of potential energy when he goes down the ramp he loses the potential energy he has and it transforms into kinetic energy

brinliest plz

6 0

3 years ago

Read 2 more answers

PLEASE HURRY!!!

lidiya [134]

Answer:

The difference between the lower mantle and the oceanic crust is first their respective locations, pressure and temperature-- the pressure and temperature increases with depth in the earth this the mantle is more hot and under great pressure than the crust.

Explanation:

8 0

3 years ago

Read 2 more answers

Paco was driving his scooter west with an initial velocity of 4 m/s. He accelerates at 0.5 m/s2 for 30 seconds. What is his fina

Kipish [7]

Answer:

19m/s

Explanation:

U= 4m/s

a = 0.5m/s²

t= 30s

V= U+at

= 4+15

=19m/s

7 0

4 years ago