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3 years ago

What do we use as a reference point to prove that we are always moving even when we are sitting down

2 answers:

4 0

Im not quite sure what you mean but do you mean us or the earth? If you mean us a reference is that our hearts are beating, we are breathing, we (may be) blinking, we are using our brains to do everything, etc.

mafiozo [28]3 years ago

3 0

This is a elementary question? in my opinion we could use stars. as the earth is in constant rotation, therefor we are technically still moving if we are sitting, just like being in a car, if that makes since

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At what angle should the axes of two polaroids be placed so as to reduce the intensity of the incident unpolarized light to 13.

prohojiy [21]

Answer:

θ = 66.90°

Explanation:

we know that

$I= \frac{I_0}{2}cos^2\theta$

I= intensity of polarized light =1

I_o= intensity of unpolarized light = 13

putting vales we get

$1= \frac{13}{2}cos^2\theta$

⇒ $\theta=cos^{-1} \sqrt{\frac{1}{6.5} }$

therefore θ = 66.90°

5 0

3 years ago

In order to increase the speed in a gear system

Ludmilka [50]

Answer:

the driving gear must be larger than the driven gear

Explanation:

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3 years ago

How do catalysts speed up reactions please help quickly

MissTica

How do catalysts speed up reactions ?

by reducing the amount of energy required for the reaction to occur. other factors such as catalysts' ability to speed a process are temperature, concentration

8 0

3 years ago

Read 2 more answers

A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What

kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let $p_{1}$ = momentum of the 1st ball

$p_{2}$ = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

$(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}$

$(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}$

Since we are dealing with identical balls, all the m terms cancel out so we are left with

$(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} + (v_{2})_{f}\cos \theta_{2}$

Putting in the numbers, we get

$1.33 = (0.750) \cos(33.30) + (v_{2})_{f} \cos \theta_{2}$

$= > (v_{2})_{f} \cos \theta_{2} = 0.703$

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

$0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}$

$= > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30) = 0.412$

Taking the ratio of the sine equation to the cosine equation, we get

$\frac{ \sin \theta _{2}}{ \cos \theta_{2} } = \tan \theta_{2} = \frac{0.412}{0.703} = 0.586$

$\theta_{2} = { \tan}^{ - 1} (0.586) = 30.4$

Solving now for $(v_{2})_{f}$ ,

$(v_{2})_{f} = \frac{0.412}{ \sin(30.4) } = 0.815 \: \frac{m}{s}$

3 0

3 years ago

A proton in a uniform electric field moves along a straight line with constant acceleration. Starting from rest it attains a vel

Sav [38]

a) The acceleration of the proton is $5.0\cdot 10^{13} m/s^2$

b) The time required to reach the given velocity is $2\cdot 10^{-8}s$

Explanation:

This is a motion at constant acceleration, so we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the proton in this problem, we have:

$v=1,000,000 m/s$ is the final velocity

$u=0$ is the initial velocity (it starts from rest)

$s = 0.01 m$ is the distance covered

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{(1,000,000)^2-0}{2(0.01)}=5.0\cdot 10^{13} m/s^2$

For this part, we can use the following suvat equation instead:

$v=u+at$

where:

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken for the velocity to change from u to v

We have here the following data:

$v=1,000,000 m/s$ is the final velocity

$u=0$ is the initial velocity (it starts from rest)

$a=5.0\cdot 10^{13} m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{1,000,000}{5.0\cdot 10^{13}}=2\cdot 10^{-8}s$

Learn more about accelerated motion here:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

5 0

3 years ago