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2 years ago

6

A 1kg object at the surface of the earth weighs 9.8N. (F=ma) Prove this by using the formula Fg = Gmm/r2. (Find the radius and t

he mass of the earth on the internet.)

1 answer:

Nadusha1986 [10]2 years ago

7 0

Answer:

$form radius \\$

g=Gm/r2

9.8=

Explanation:

$9.8 = (6.67 \times 10 ^{ - 2} \times 1kg) \div r \ r = 6.806 \times 10 ^{ - 12} \\ r = 2.6 \times 10 ^{ - 6}$

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Suppose you pluck a string on a guitar and it produces the note A at a frequency of 440 Hz. Now you press your finger down on th

MAVERICK [17]

Answer:

D. 550 Hz

Explanation:

Frequency for string is given by

$f_{} =\frac{v}{2L}$

where v is the speed of the wave, L = m is the length

Frequency is inversely proportional to the Length of the string

$f \alpha \frac{1}{L} \\$

from above relation we can write

$\frac{f_{1}}{f_{2}} =\frac{L_{2}}{L_{1}}$

The newly shortened string has 4/5 the length of the full string .i,e

$\frac{440}{f_{2}} =\frac{\frac{4}{5}L_{1} }{L_{1}}\\f_{2}=\frac{440\times5}{4} \\f_{2}=550 Hz$

Hence option D is correct

8 0

3 years ago

a diver is standing on the diving board. the diver has a mass of 85.0kg and has a gravitational potential emergy of 2.50x103 j s

amid [387]

Answer:

<h2>3.00m high</h2>

Explanation:

Gravitational potential energy of the diver is expressed mathematically as

P = mgh

m = mass of the diver (in kg)

g = acceleration due to gravity (in m/s²)

h = the height of the diver above the water surface.

Given m = 85.0kg, g = 9,8m/s² and P = 2.50x10³ Joules, wecan determine how high the diver is above the water surface bu substituting the given values into the formula;

h = P/mg

h = 2.50x10³/85*9.8

h = 2500/833

h ≈ 3.00m

The diver is 3.00m above the water surface.

5 0

3 years ago

Tpy6a [65]

The answer is D) Gravity is dependent on mass and strong nuclear force is not.

pls mark me brainliest

5 0

3 years ago

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According to Kepler's Third Law, a solar-system planet that has an orbital period of 8 years would have an orbital radius of abo

ELEN [110]

Answer:

Orbital period, T = 1.42 years

Explanation:

It is given that,

Orbital period of a solar system planet, $T=8\ years=2.52\times 10^{8}\ s$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$r^3=\dfrac{T^2GM}{4\pi^2}$

$r^3=\dfrac{(2.52\times 10^{8})^2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}}{4\pi^2}$

$r^3=2.134\times 10^{35}$

$r=5.975\times 10^{11}\ m$

r = 1.42 AU

So, the solar-system planet that has an orbital period of 8 years would have an orbital radius of about 1.42 AU.

6 0

2 years ago

When force is ---- to the direction of motion no work is?

xxTIMURxx [149]

As work is a vectorial product it is directly dependant on displacement and force

When force is zero, but displacement occurs ----> no work is done

Like-wise:
If there is force but no displacement --> work is also not done

3 0

3 years ago

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