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2 years ago

7

Help on 9 and 10!!!!!! I’ll give 5 starts

1 answer:

julia-pushkina [17]2 years ago

4 0

Answer: part a is e

part b a

Explanation:

I HOPE THIS HELPS

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A helicopter's speed increases

sergij07 [2.7K]

Answer:

the helis speed increased its speed by 35 m/s. Unless you are asking me what was the speed per second. if you were asking me for the speed per second then that would be a different answer

Explanation:

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3 years ago

Describe two reasons why an alpha particle is less penetrating than a beta or gamma particle.

uysha [10]

Alpha particles travel through the air they collide with oxygen and nitrogen molecules. While they collide with these molecules, they lose some energy until all energy are used up and they are absorbed. These particles can be absorbed by a sheet of paper or by the air. On the other hand, beta particles and gamma particles move faster than the alpha particles and are poor at ionizing atoms or molecules thus it takes more of the material to be able to absorb these particles.

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3 years ago

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A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They

soldier1979 [14.2K]

Answer:

a. $\thta=71^{\circ}$

b. $v_f=3.78 m/s$

Explanation:

We are given that

$m_1=53 kg$

$v_1=2.9 m/s$

$m_2= 72 kg$

$v_2=6.2 m/s$

a.We have to find the angle

$\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}$

$\theta=71^{\circ}$

b. We have to find the speed $v_f$

According to law of conservation of momentum

$m_1v_1=(m_1+m_2)v_fcos\theta$

$53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f$

$v_f=\frac{153.7}{40.7}=3.78 m/s$

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3 years ago

A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with

Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

$V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)$

Charge density of the shell is,

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}$

Here, $R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q$

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}$

B)

The electric field is $E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}$

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

$E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)$

substitute $\rho=\frac{-3Q}{28\piR^3}$

$E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}$

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

8 0

3 years ago

We can determine the __________ of a wave when given the frequency and the wavelength. (2 points)

Olin [163]

The correct answer to the problemis velocity. I did the test.

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2 years ago

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