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3 years ago

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