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Andreas93 [3]
3 years ago
6

The distance from a crest to a trough is called the wave length

Physics
1 answer:
V125BC [204]3 years ago
5 0

Answer:

Explanation:

The highest surface part of a wave is called the crest, and the lowest part is the trough. The vertical distance between the crest and the trough is the wave height. The horizontal distance between two adjacent crests or troughs is known as the wavelength.

I hope it's helpful!

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What help in reversing direction of current of current​
Novay_Z [31]

Answer:

<em>To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection.  </em>

Explanation:

For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.

3 0
4 years ago
. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa
lozanna [386]

Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

<u>Vf = 128 ft/s</u>

<u></u>

(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

<u>K.E = 122.8 Btu</u>

3 0
4 years ago
A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with
Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = \dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2

              = \dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2

              = 31 J

Final KE = \dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0
4 years ago
An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic
Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So 90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with a=2m/sec^2 for 7 sec

So distance traveled by car in these 7 sec

S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so 224=\left ( 1+7 \right )v

v = 28 m/sec

4 0
4 years ago
Read 2 more answers
a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car​
uysha [10]

Answer:

= 201.53 meters

Explanation:

A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?

Use the formula  d = \frac{at^{2} }{2} ,

where d is the distance, t is the time and "a" is the acceleration.
d=\frac{9*54*6*5^{2} }{2} = 201.53 m

3 0
2 years ago
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