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Setler [38]
3 years ago
9

A car travels a distance of 60 km in 2 hours. what is its speed in kilometers per hour?

Physics
1 answer:
White raven [17]3 years ago
5 0

Answer:

30 kilometers per hour.

Explanation:

we know this by dividing 60 by 2

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The National Aeronautics and Space Administration (NASA) studies the physiological effects of large accelerations on astronauts.
m_a_m_a [10]

Answer:

v = 23.95 m/s

Explanation:

As we know that when astronaut is revolving in circular path then the acceleration of the astronaut is due to centripetal acceleration

so it is given as

a_c = \frac{v^2}{R}

here we know that

a_c = 4.50 g

also we know that

R = 13 m

now we have

4.50 \times 9.81 = \frac{v^2}{13}

v = 23.95 m/s

3 0
4 years ago
Humpback whales are known to produce a collection of elaborate and repeating sounds with frequencies starting at 20 Hz. The soun
jasenka [17]

Answer:

70 m.

Explanation:

Given,

Frequency, f = 20 HZ

speed of sound, v = 1400 m/s

wavelength of the waves = ?

we know,

v = f λ

\lambda= \dfrac{v}{f}

\lambda= \dfrac{1400}{20}

\lambda=70\ m

Hence, the wavelength of the wave is equal to 70 m.

8 0
4 years ago
The 6 kg block is then released and accelerates to the right, toward the 4 kg block. The surface is rough and the coefficient of
KatRina [158]

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

U = \dfrac{1}{2}kx^2

U = \dfrac{1}{2}\times 350 \times 0.5^2

   U = 43.75 J

using conservation of energy

 KE = U - f.d

where f is the frictional force acting

\dfrac{1}{2}mv^2 = 43.75- \mu m g d

\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5

v= \sqrt{11.643}

       v =3.41 m/s

4 0
4 years ago
A train is traveling at a speed of 75km/hr. The brakes are applied to produce a uniform acceleration of -0.5 m/s2.Find how far t
NNADVOKAT [17]

Answer:

your answer would be 21m

Explanation:

7 0
3 years ago
A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns
Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

f'=\frac{V+V_{o}}{V-V_{s}} f  (1)

Where:

f=250,000 Hz is the actual frequency of the sound wave

f'=240,000 Hz is the "observed" frequency

V=340 m/s is the velocity of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which is the balloon

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}  (2)

V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}  (3)

V_{s}=-14.16 m/s (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time (t=4.8 s) given as data, we can find the distance:

d=V_{s}t (5)

d=(14.16 m/s)(4.8 s) (6)

Finally:

d=68 m (8) This is the distance of the balloon from the radar station

6 0
3 years ago
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