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irakobra [83]
3 years ago
12

In a transverse wave that travels through a medium, the molecules of the medium vibrate

Physics
1 answer:
LenaWriter [7]3 years ago
4 0
<span>So we wan't to know how the molecules of the medium vibrate when a transverse wave propagates trough that medium. A transverse wave is a wave whos oscilations are perpendicular to the direction of it's propagation. So the molecules in a medium will vibrate perpendicular to the direction of propagation of the wave. So if the wave is moving in the x-direction, the molecules will vibrate in the y-z plane. The correct answer is A. at right angles to the direction in which the wave propagates. </span>
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A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position
Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

v_x= \frac{dx}{dt}

Write an equation for x.

dx=v_xdt\\ x=\int v_xdt

Substitute the equation for vₓ =2t² in the integral.

x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m

Substitute 2.0s for t.

x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m

The position of the particle at t =2.0 s is <u>6.4m</u>




5 0
3 years ago
A car has uniformly accelerated from rest to a speed of 25m/s after traveling 75m. What is its acceleration in m/s^2
Roman55 [17]
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3 years ago
Please help.. I'm desperate.
rosijanka [135]

Answer:

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Explanation:

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3 years ago
a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on
alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

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3 years ago
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Arte-miy333 [17]

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I'm pretty sure it means like getting abs and working out your stomach.

Explanation:

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2 years ago
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