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3 years ago

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Calculate the force of gravity between two planets with masses 5 x 1023 kg and 1 x 1022 kg when the distance

1 answer:

ipn [44]3 years ago

6 0

Answer:

68.1N

Explanation:

Check attachment

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How are the sun, the moon, and Earth related during a solar eclipse?

german

B. The moon is located between the Sun and Earth

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3 years ago

Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov

Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

$K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}$

Final kinetic energy is :

$K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}$

Now, fraction of initial kinetic energy loss is :

$Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25$

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0

3 years ago

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Serhud [2]

A baseball will curve better on the flat plain if it is higher than sea level but low elevation.

Hope this helped!

4 0

3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

3 years ago

A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a

andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

$sin\theta = 1.22\frac{\lambda}{d}$

Here,

$\lambda$ = Wavelength

d= Diameter of aperture

$\theta$ = Angular resolution or diffraction angle

Our values are given as,

$\theta = 11\°$

The frequency of the sound is $f = 9100 Hz$

The speed of the sound is $v = 343 m/s$

The wavelength of the sound is

$\lambda = \frac{v}{f}$

Here,

v = Velocity of the wave

f = Frequency

Replacing,

$\lambda = \frac{(343 m/s)}{(9100 Hz)}$

$\lambda = 0.0377 m$

The diffraction condition is then,

$sin\theta = 1.22\frac{\lambda}{d}$

Replacing,

$sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}$

d = 0.24 m

Therefore the diameter should be 0.24m

6 0

3 years ago