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3 years ago

10

Electric Charge (guided inquiry)

1 answer:

Sedaia [141]3 years ago

8 0

Answer:

It looks like you have some writing to do ,(-:

Explanation:

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Two cars are traveling on a desert road. After 5.0 seconds they are side by side at the next telephone pole.The distance between

stira [4]

The question looks incomplete, but according to the information given above seem like they have <span>identical journeys.
</span>a. the displacement of car A - <span> 65.5 m
</span>b. the displacement of car B - <span>65.5 m
c. average velocir</span>y of A $65.5 / 5 = 13.1 m/s$
d. the average velocity of car B has the same.

8 0

3 years ago

Which of the following best describes what muscles do.

kow [346]

I think b support body weight

8 0

3 years ago

Read 2 more answers

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

a

The mass is $m_2 =21.75*10^{-27} \ kg$

b

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

So

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

So

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

7 0

3 years ago

A fan blade, initially at rest, rotates with a constant acceleration of 0.029 rad/s2. What is the time interval required for it

LenKa [72]

Answer:

The time interval is $t = 21.30 \ s$

Explanation:

From the question we are told that

The constant acceleration is $\alpha = 0.029 \ rad / s^2$

The displacement is $\theta = 6.58 \ rad$

According to the second equation of motion we have that

$\theta = w_i* t + \frac{1}{2} * \alpha t^2$

given that the blade started from rest

$w_i$ which is the initial angular velocity is 0

So

$\theta = 0 + \frac{1}{2} * \alpha t^2$

=> $t = \sqrt{ \frac{2 * \theta }{\alpha } }$

substituting values

=> $t = \sqrt{ \frac{2 * 6.58 }{0.029 } }$

=> $t = 21.30 \ s$

6 0

3 years ago

6. What IMFA holds the molecules of water together?

Goryan [66]

The answer is Polar covalent. btw, this is chemistry work. not physics

5 0

3 years ago