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3 years ago

15

A chipmunk with a mass of 2 kg sits on a branch that is 10 m off the ground.

1 answer:

vlada-n [284]3 years ago

3 0

I hope I'm not too late.

GPE = mass * gravity * height

GPE = 2 kg * 9.8 m/s * 10

GPE = 196 Joules

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Can a nickel be attracted to a magnet?

zhuklara [117]

Yes... This is a question google could answer. Just Saying

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3 years ago

How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

Tresset [83]

Answer:

<em>0.97c</em>

<em></em>

Explanation:

From the relativistic equation for length contraction, we have

$l$ = $l_{0}\sqrt{1 - \beta }$

where

$l$ is the final length of the object

$l_{0}$ is the original length of the object before contraction

β = $v^{2} /c^2$

where v is the speed of the object

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

The equation can be re-written as

$l$ / $l_{0}$ = $\sqrt{1 - \beta }$

For the length to contract to one-fourth of the proper length, then

$l$ / $l_{0}$ = 1/4

substituting into the equation, we'll have

1/4 = $\sqrt{1 - \beta }$

substituting for β, we'll have

1/4 = $\sqrt{1 - v^2/c^2 }$

squaring both side of the equation, we'll have

1/16 = 1 - $v^2/c^2$

$v^2/c^2$ = 1 - 1/16

$v^2/c^2$ = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = <em>0.97c</em>

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3 years ago

Parents have a major role in observing how young children grow at home. Create and discuss simple developmental monitoring tips

Rudik [331]

Answer:

by observation of their children

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2 years ago

Please answer! thank you!

lesantik [10]

Answer:

Maybe A.... I'm not sure

Explanation:

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3 years ago

Read 2 more answers

An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line t

ki77a [65]

Explanation:

We will use the equations of constant acceleration to find out $a_{x}$ and time t.

As we know that the initial speed is zero. So

(a)

$v_{0x} = 0$

$x - x_{o} = 1.25$ × $10^{-2}$ m

$v_{x} = 3.3$ × $10^{6}$ m/s

$v^{2} _{x} = v^{2} _{x_{o} } + 2a_{x} (x - x_{o} )$

$a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }$

= $\frac{(3.3 * 10^{6})^{2} - 0 }{2(1.25 * 10^{-2}) }$

= 4.356× $10^{14}$ m/s²

(b)

$v_{x} = v_{ox} + a_{x}t$

$t = v_{x} - vo_{x}/a_{x}$

$t = \frac{3.00 * 10^{6} }{4.356*10^{14} }$ = 6.8870× $10^{-9}$ s

(c)

Σ $F_{x} = ma_{x}$

= (9.11× $10^{-31}$ )(4.356× $10^{14}$ m/s²)

= 3.968× $10^{-16}$ N

6 0

3 years ago