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erastovalidia [21]
3 years ago
13

Now set the tension to low and wiggle the wrench to create more waves. Can you explain how moving the first point on the string,

the one closest to the wrench, affects the next point on the spring? How does this fit with your understanding of the force of a stretched spring?
Physics
2 answers:
IrinaK [193]3 years ago
5 0

Answer:

As the first particle travels upward, it pulls on the next particle, which follows the first particle upward. Then next in line is pulled up in turn, and so on. The motion of each particle follows the one before it, either up or down, with a slight lag in time. This succession of particles moving up or down travels along the string as a wave. Eventually, it pulls every particle along the string up and down in series.

Explanation:

Hatshy [7]3 years ago
4 0

Answer:

When the string moves, it creates a very small change in the distance to the next point, th

Explanation:

When the string moves, it creates a very small change in the distance to the next point, this generates a restoring force that tends to push the string back, this small disturbance propagates along the string and is what creates the pulse.

This is similar to what happens when a spring is stretched and a restoring force is generated shaved by the law of shortening.

            F = k Dx

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The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
nataly862011 [7]

Answer with Explanation:

Let  rest mass m_0 at point P  at  distance x from center of the planet, along a line connecting the centers of planet and the moon.

Mass of moon=m

Distance between the center of moon and center of planet=D

Mass of planet=M

We are given that net force on an object will be zero

a.We have to derive an expression for x in terms of m, M and D.

We know that gravitational force=\frac{GmM}{r^2}

Distance of P from moon=D-x

F_m=Force applied on rest mass due to m

F_m=Force on rest mass due to mas M

F_M=F_m because net force is equal to 0.

F_m=F_M

\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}

\frac{m}{(D-x)^2}=\frac{M}{x^2}

\frac{x^2}{(D-x)^2}=\frac{M}{m}

\frac{x}{D-x}=\sqrt{\frac{M}{m}}

Let R=\sqrt{\frac{M}{m}}

Then, \frac{x}{D-x}=R

x=DR-xR

x+xR=DR

x(1+R)=DR

x=\frac{DR}{1+R}

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x=\frac{2}{3}D

Net force is zero

F_m=F_M

\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}

\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}

\frac{M}{m}=4

Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

8 0
4 years ago
Part 1: What is your Reaction Time?
nlexa [21]

Answer:

1. A ruler is a common instrument used for measuring the length of small objects

2. The acceleration due to gravity at the surface of Earth is about 9.8 metres per second per second.

8 0
3 years ago
In a single-slit diffraction experiment, a coherent light source illuminates a slit in a barrier, and the resulting pattern is p
UkoKoshka [18]

Answer:

hsjdsdddwqdqwdd

Explanation:

4 0
3 years ago
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0
3 years ago
A ladder placed up against a wall is sliding down. The distance between the top of the ladder and the foot of the wall is decrea
Kitty [74]

Answer:

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance  h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L²   ..................1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt     .............2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223  (-9)

so

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

3 0
3 years ago
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