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photoshop1234 [79]
3 years ago
13

An astronaut takes an iPod onto the space shuttle. An identical iPod remains on Earth. Which statement about the pull of Earth's

gravity on the iPods is true?

Physics
2 answers:
maria [59]3 years ago
5 0
The answer is actually B 
puteri [66]3 years ago
3 0

Answer:

B is the answer because if you throw a phone in space it will hit the ground way slower than if you threw it on earth.

earth pulls the phone back down faster because the earth's gravity is stronger!

Explanation:

The answer can't be A because you have less gravity in space.

C is incorrect because earth's gravity forces us down, but in space you can float

D is incorrect for the same reason as A

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Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron
TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m

Next, you calculate the wavelength of the sound waves by using the following formula:

\lambda=\frac{v_s}{f}

vs: speed of sound =  343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

\lambda=\frac{343m/s}{400Hz}=0.8575m

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

\Delta d=3.827m-2.10m=1.727m

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2

Then, there will be a constructive interference, and the sound who the observer hears is loud.

5 0
3 years ago
It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s 2 . Find the magnitude of acce
shepuryov [24]

This Question is not complete

Complete Question:

a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.

Answer in units of m/s2

b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under

these new conditions.

Answer in units of m/s2

Answer:

a. 2.875m/s²

b. 3.172m/s²

Explanation:

a. The formula for centripetal acceleration = (speed²) ÷ radius

Centripetal acceleration = (5.7m/s)²÷ 11.3m

Centripetal acceleration = 2.875m/s²

b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.

Centripetal acceleration ( acceleration x) = 2.875m/s²

Increase in the speed rate ( acceleration n) = 1.34m/s²

Magnitude of acceleration = √a²ₓ + a²ₙ

=√( 2.875m/s²)²+ (1.34m/s²)²

= √ 10.06m/s²

= 3.172m/s²

3 0
4 years ago
Read 2 more answers
A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b)
kondaur [170]

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s

Initial pop up velocity is 35.87 m/s

a)

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s

It took 3.65 seconds to reach this height

6 0
3 years ago
Basketball player Darrell Griffith is on record as
Gre4nikov [31]

Explanation:

1.

We use the equation

h = \frac{gt^2}{2}, where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

2.

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

5 0
3 years ago
List six forms of electromagnetic radiation from the shortest waves(highest energy) to the longest waves (lowest energy)
romanna [79]
That's the reverse of RIVUXG
so your answer is
gamma rays
x rays
ultraviolet light
visible light
infrared
radio waves
8 0
3 years ago
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