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2 years ago

Can anyone tell me the ans of this question with steps of the solution

Physics

2 answers:

ArbitrLikvidat [17]2 years ago

4 0

Answer:

64m

Explanation:

100sec=160×0.8

100sec=128m

50sec=less

=50/100×128m

=0.5×128

=64m

gavmur [86]2 years ago

3 0

Answer:

Explanation:

100s:160 paces

50s: 80 paces

80*0.80= 64

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What is the energy of a photon with a frequency of 5 x 10^14 Hz?

zloy xaker [14]

Answer:

$3.315 x 10^-19 J$

Explanation:

Using Eisten's formula, energy is a product of Plank's constant and photon frequency

E=hf

Where f represent frequency, h is Plank's constant and E is energy.

Substituting frequency with 5 x 10^14 Hz and

Planck's constant is 6.63 x 10^-34 J•s.

$E=6.63 x 10^{-34} *5 x 10^{14}=3.315x10^{-19} J$

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3 years ago

What are two reasons why density is a useful physical property for identifying substances?

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Some substances look alike and the density can help you determine which is which. It also helps to tell which is heavier than the other.

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3 years ago

What is point c called?

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2 years ago

Read 2 more answers

you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?

laiz [17]

At the same speed because it will slow down as it approaches the peak then speed up as it goes down again

it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something

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2 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

2 years ago