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steposvetlana [31]

3 years ago

11

You’re squeezing a springy rubber ball in your hand. If you push inward on it with a force of 1 N, it dents inward 2 mm. How far

must you dent it before it pushes outward with a force of 5 N?

1 answer:

creativ13 [48]3 years ago

8 0

Answer:

10mm

Explanation:

According to Hooke's law which states that "the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Direct proportionality there means, increase/decrease in the force leads to increase/decrease in extension.

Mathematically, F = ke where;

F is the applied force

k is the elastic constant

e is the extension

from the formula k = F/e

k = F1/e1 = F2/e2

Given force of 1N indents the spring inwards by 2mm, this means force of 1N generates extension of 2mm

Let F1 = 1N e1 = 2mm

The extension that will be produced If force of 5N is applied to the string is what we are looking for. Therefore F2 = 5N; e2= ?

Substituting this values in the formula above we have

1/2=5/e2

Cross multiplying;

e2 = 10mm

This shows that we must have dent it by 10mm before it pushes outwards by a 5N force

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A colony of troglodytes has been in a lengthy feud with

Troyanec [42]

Answer:

h = 181.73 m

Explanation:

given,

distance between the neighbors = 42 m

speed of the rock rolling = 6.9 m/s

vertical velocity of ball = 0 m/s

height of the cliff = ?

time taken by the ball to travel 42 m

d = s x t

s is the horizontal speed of the ball equal to 6.9 m/s

$t =\dfrac{d}{s}$

$t =\dfrac{42}{6.9}$

t = 6.09 s

same time will be taken by the ball to travel vertical distance

Using equation of motion

$h = u t + \dfrac{1}{2}gt^2$

$h = 0+ \dfrac{1}{2}gt^2$

$h = \dfrac{1}{2}\times 9.8 \times 6.09^2$

h = 181.73 m

7 0

3 years ago

Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by

mylen [45]

Answer:

<em>I must travel with a speed of 2.97 x 10^8 m/s</em>

Explanation:

Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months

Time that elapses on the spacecraft = 1 year

On earth the people have advanced 120 yrs

According to relativity, the time contraction on the spacecraft is gotten from

$t$ = $t_{0} /\sqrt{1 - \beta ^{2} }$

where

$t$ is the time that elapses on the spacecraft = 120 years

$t_{0}$ = time here on Earth = 1 year

$\beta$ is the ratio v/c

where

v is the speed of the spacecraft = ?

c is the speed of light = 3 x 10^8 m/s

substituting values, we have

120 = 1/ $\sqrt{1 - \beta ^{2} }$

squaring both sides of the equation, we have

14400 = 1/ $(1 - \beta ^{2} )$

14400 - 14400 $\beta ^{2}$ = 1

14400 - 1 = 14400 $\beta ^{2}$

14399 = 14400 $\beta ^{2}$

$\beta ^{2}$ = 14399/14400 = 0.99

$\beta = \sqrt{0.99}$ = 0.99

substitute β = v/c

v/c = 0.99

but c = 3 x 10^8 m/s

v = 0.99c = 0.99 x 3 x 10^8 = <em>2.97 x 10^8 m/s</em>

6 0

3 years ago

Two satellites, A and B are in different circular orbits

jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

6 0

3 years ago

PLEASE ANSWER ASAP

neonofarm [45]

Answer:

2.55sec

Explanation:

time = distance/speed = (87 m)/(34 m/s) = 2.55sec

6 0

2 years ago

You are listening to the radio when one of your favorite songs comes on, so you turn up the volume. If you managed to increase t

andrew-mc [135]

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

$\beta_{dB} = 10log_{10} \frac{I}{I_0}$

Where,

I = Acoustic intensity in linear scale

$I_0$ = Hearing threshold

The value in decibels is 17dB, then

$17dB = 10log_{10} \frac{I}{I_0}$

Using properties of logarithms we have,

$\frac{17}{10} = log_{10} \frac{I}{I_0}$

$log_{10} \frac{I}{I_0} = 1.7$

$\frac{I}{I_0} = 10^{1.7}$

$\frac{I}{I_0} = 50.12 W/m^2$

Therefore the factor that the intensity of the sound was $50.12W/m^2$

5 0

3 years ago