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4 years ago

6

If you were in a freely falling elevator and you dropped a pencil. it would hover in front of you. Is there a force of gravityac

ting on the pencil? Defend your answer.

1 answer:

11111nata11111 [884]4 years ago

7 0

Answer:Yes

Explanation:

Yes, there will be a force of gravity acting on the Pencil but it will hover in front of the person because lift would itself having free fall.

Free is the state of body where gravity is the only force acting on the body while it is falling from a height.

According to the theory of relativity, a body under free fall has no force acting on it.

Anything inside the lift will be experiencing the free fall as they are falling under the influence of gravity completely.

You might be interested in

Which is the best description of a molecule?

Mazyrski [523]

Answer:

A molecule of an element is composed of exactly two types of atoms.

Explanation:

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3 years ago

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Determine the average velocity and average speed for the following situation. A dog runs west for 35 meters then turns east for

mart [117]

Answer:

0.882 m/s average velocity and 1.71 m/s average speed

Explanation:

The dog travels a total of 35 m west and 110 m east.

110-35 = 75 m east of the starting position. Since velocity is a vector you must consider its first and final position and not the total distance traveled.

75 m / 85 s = 0.882 m/s average velocity

Speed is not concerned with direction so we instead add the total distance traveled which is 35+110 = 145 m. We then perform the same operation as before and divide by the time it took to run this distance.

145 m / 85 s = 1.71 m/s average speed

7 0

3 years ago

An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise

Viefleur [7K]

Answer:

The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

$\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}$

$\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}$

$f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)$

$\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)$

$\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

$v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

Substitute the given values in the formula,

$v_{s}=343+\frac{262}{271}(343-0)$

$v_{s}=343+0.966(343)$

$v_{s}=343-331.33$

$v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Therefore, The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

4 0

3 years ago

A simple pendulum is mounted in an elevator. What happens to the period of the pendulum (does it increase, decrease, or remain t

Semenov [28]

Answer:

Explanation:

Time period of Simple pendulum is given by

$T=2\pi\sqrt{\frac{L}{g_{effective}}}$

where L=length of Pendulum

g=acceleration due to gravity

(a)When elevator is accelerating upward with $5 m/s^2$

effective g will be g+a=9.8+5

therefore time period will decrease

(b)When elevator is moving upward with 5 m/s

effective g will be g as there is no added acceleration

therefore time period will remain same

(c)When elevator is accelerating downward with $5 m/s^2$

effective g will be g-a=9.8-5

therefore time period will increase

(d)When elevator is accelerating downward with $9.8 m/s^2$

effective g will be g-9.8=0

therefore time period will become infinite

4 0

3 years ago

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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

$a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev$

$d= 155 rev = 155(2\pi ) = 310\pi rad$

$a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi ) = 4.0\pi \frac{rev}{s^{2} }$

$d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449$

$w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}$

Now the angular velocity is the blade speed so:

$V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}$

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

$t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s$

Now is the same time the piece travel horizontally

$L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m$

blade piece travels HORIZONTALLY = (24.5)(0.397) = 9.73 m ANS

6 0

4 years ago