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Feliz [49]
2 years ago
5

A transverse wave has a frequency of 200 Hz with a wavelength of 1.0 m. Determine the speed

Physics
1 answer:
Lana71 [14]2 years ago
7 0

Answer:

200 m\ s Ans .....

Explanation:

Data:

f = 200 Hz

w = 1.0 m

v = ?

Formula:

v = f w

Solution:

v = ( 200)(1.0)

v = 200 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
atroni [7]

Answer:

The average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V

So, the average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

6 0
3 years ago
Read 2 more answers
Match the lithification processes.
butalik [34]

Explanation:

Recrystallization: contact pressure causing grains to "fuse" together

Cementation : precipitation of bonding agents between grains

Compaction : increase in density due to weight of overburden

Lithification is the process by which sediments are converted into sedimentary rocks. During this process, recrystallication, compaction and cementation of mineral grains occur.

The process starts with the compaction of sediments. The over burden weight of new sediments in the basin adds to the one originally deposited. This compresses the sediment. The volume of reduced and the density increases.

Recrystallization follows suit as the contact pressure of grains makes them fuse together. It is more like reworking of sediments. In this process, cementing materials can precipitate and cause sediments to be more fused together.

This is why most sediment are made up of clasts in a matrix of cementing materials.

learn more:

sedimentary rocks brainly.com/question/9131992

#learnwithBrainly

4 0
2 years ago
The gravitational force between two objects (mass1 = 10kg, mass2 = 6kg) is measured when the objects are 12 centimeters apart. I
Yanka [14]
Since the new distance is 3 times the old distance,
the new force is (1/3²) = 1/9th of the old force.

That's kind-of Choice-D, but I really don't like the way choice-D is worded.
"9 times smaller" is really pretty meaningless.  
8 0
3 years ago
A force of 42N is needed to start a box sliding across the floor. The weight of the box is 55N.
Bezzdna [24]
I think the right answer is c
4 0
3 years ago
Read 2 more answers
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
3 years ago
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