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givi [52]
3 years ago
9

A space station of diameter 20.0 meters is turning about its axis to simulate gravity at its center rim. How fast must it rotate

to produce an outer rim acceleration of 9.80 m/s^2 ?
Physics
1 answer:
bezimeni [28]3 years ago
4 0

Answer:

9.89 m/s

Explanation:

d = diameter of the space station = 20.0 m

r = radius of the space station

radius of the space station is given as

r = (0.5) d

r = (0.5) (20.0)

r = 10 m

a = acceleration produced at outer rim = 9.80 m/s²

v = speed at which it rotates

acceleration is given as

a = \frac{v^{2}}{r}

9.80 = \frac{v^{2}}{10}

v = 9.89 m/s

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What is speed of wave in terms of frequency and wavelength??​
ratelena [41]

Answer:

In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form, If the crest of an ocean wave moves a distance of 20 meters in 10 seconds, then the speed of the ocean wave is 2.0 m/s.

Speed = Wavelength x Wave Frequency. In this equation, wavelength is measured in meters and frequency is measured in hertz (Hz), or number of waves per second. Therefore, wave speed is given in meters per second, which is the SI unit for speed.

Explanation:

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3 0
2 years ago
A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor r
Anika [276]

Answer:

156.8 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 10 kg

Height (h) = 8 m

Time (t) = 5 s

Power (P) =?

Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:

Mass (m) = 10 kg

Height (h) = 8 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 10 × 9. 8 × 8

E = 784 J

Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:

Time (t) = 5 s

Energy (E) = 784 J

Power (P) =?

P = E/t

P = 784 / 5

P = 156.8 Watts

Therefore, the power output of the motor is 156.8 Watts

7 0
3 years ago
A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w
aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂= 0.6kg

4 0
3 years ago
A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed?
Sunny_sXe [5.5K]

Answer:

v=500 m/s

<h3>Solution:</h3>

v=(10m)(50/s)

5 0
2 years ago
Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109
r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

E=\Delta mc^2\\\Delta m=\frac{E}{c^2}(1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J

Hence, by replacing in the equation (1) you have  (c=3*10^{8}m/s)

\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg

HOPE THIS HELPS!!

3 0
3 years ago
Read 2 more answers
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