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saw5 [17]
3 years ago
6

A wave with a frequency of 14Hz has a wavelength of 3 meters. At what speed will this wave travel

Physics
1 answer:
3241004551 [841]3 years ago
7 0

Answer:

frequency = 14 Hz

wavelength = 3 m

speed = ?

speed = frequency

x wavelength

speed =    14  x  3speed =     42 meter per second

Explanation:

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Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.9 cm and a current of 12 A. The bigger lo
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Explanation:

Below is an attachment containing the solution.

4 0
3 years ago
A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
3 years ago
a 16.0 kg cart is being pulled by a 95.4 N force to the right and a 36.0 N force to the left. What is the acceleration of the ca
Inga [223]

Answer:

The cart's acceleration is \approx 3.71\,\,\frac{m}{s^2}

Explanation:

Let's start by finding the net force acting on the cart, and then find its acceleration using Newton's 2nd Law.

Net force = 95.4 N -36.0 N = 59.4 N

Now, since we know the cart's mass, we can use Newton's 2nd Law to find the cart's acceleration:

F=m\,*\,a\\a=\frac{F}{m} \\a=\frac{59.4}{16} \,\,\frac{m}{s^2} \\a\approx 3.71\,\,\frac{m}{s^2}

8 0
3 years ago
A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin
gogolik [260]

Answer:

a) 23 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f}  (1)

  • The initial momentum p₀, can be written as follows:

       p_{o} =  m_{1}  * v_{1o} + m_{2}* v_{2o} =   6.0 kg * 5.0 m/s + 2.0 kg * v_{2o}  (2)

  • The final momentum pf, can be written as follows:

        p_{f} = (m_{1} + m_{2} )* v_{f}  = 8.0 kg* (-2.0 m/s)  (3)

  • Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg}  = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s  (4)

  • This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.
6 0
3 years ago
The light from many stars can be seen from earth. but there is a time delay between the time the light is emitted from the star
mrs_skeptik [129]
Stars are located at a distance which are measured in terms of light years. Light year is an Astronomical unit used to measure distance between distant Celestial bodies. 
1 light year = 9460730472580<span>800 metres 
But no star is located at a distance of 1 light year. Some stars are located at millions of light years and light travels ~ 3 x 10</span>⁸ m/s. Thus light takes time to reach our atmosphere. 
4 0
4 years ago
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