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tatyana61 [14]
3 years ago
7

680 hertz 0.5 meters whats the speed of the sound of the wave

Physics
1 answer:
Novay_Z [31]3 years ago
3 0
The velocity of any wave is given by:

v=f \lambda

In which f is the wave's frequency (measured in hertz) and \lambda is the wave's wavelength (measured in meters).

So, for this wave:
v=(680)(.5)=340 \frac{m}{s}
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A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran
marishachu [46]
<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train  = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train  =  4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

8 0
3 years ago
Why is newton's gravitational force called universal gravitational force​
Lelechka [254]
Newton’s law of gravity is considered “universal” because it is believed to be applicable to the entire Universe (to a good approximation). Gravity does not just pull an apple from a tree, but also pulls the Moon to the Earth, the Earth to the Sun, and so on. Gravity is a fundamental force of nature.
6 0
3 years ago
Read 2 more answers
A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st
sp2606 [1]

Answer:

A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2  ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck :       m1 = 2000 Kg           v1 = 5 m/s                u1 = 0

car    :        m2 = 1000 kg           v2 = 0                      u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = \frac{2000}{1000} x 5

    = 10 m/s

Hence after collision, car moves at a velocity of 10 m/s

3 0
3 years ago
The earth spins on its axis once a day and orbits the sun once a year (365 1/4 days). Determine the average angular velocity (in
lubasha [3.4K]

Answer:

Given that

The earth spins on its axis once a day and orbits the sun once a year (365 1/4 days)

a)

When earth spins on its axis

We know that earth take 1 day to complete one revolution around its own axis.

T= 1 day = 24 hr = 24 x 3600 s

T=86400 s

We know that

T=2π/ω

ω= 2π/T

ω= 2π/86400

ω=7.27 x 10⁻5 rad/s

b)

When earth revolve around earth

T =365 1/4 days = 365.25 days

T= 365.24 x 86400 s

T=31557600

We know that

T=2π/ω

ω= 2π/T

ω= 2π/31557600

ω=1.99 x 10⁻⁷ rad/s

8 0
2 years ago
Read 2 more answers
A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?
vesna_86 [32]

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

v^2=u^2+2as

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

v=u+at

where symbols have the usual meaning

Applying the given values we get

t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds

4 0
3 years ago
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