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levacccp [35]
3 years ago
5

ly charged particles are held 24 x 103m apart and then released from rest. The initial acceleration of the first particle is obs

erved to be 7.0 m/s2 and that of the second to be 9.0 m/s2. The mass of the first particle is 6.3 x 107 kg. (a) What is the mass of the second particle? kg (b) What is the magnitude of the charge of each particle?
Physics
1 answer:
inessss [21]3 years ago
3 0

Answer:

Part a)

m_2 = 4.9 \times 10^7 kg

Part b)

q_1 = q_2 = 5312.6 C

Explanation:

Part a)

As we know that both charge particles will exert equal and opposite force on each other

so here the force on both the charges will be equal in magnitude

so we will have

F = m_1a_1 = m_2a_2

here we have

6.3 \times 10^7(7) = m_2(9)

now we have

m_2 = 4.9 \times 10^7 kg

Part b)

Now for the force between two charges we can say

F = \frac{kq_1q_2}{r^2}

now we have

(6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}

now we have

q_1 = q_2 = 5312.6 C

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2. A 4.0 kg magnetic toy car traveling at 3.0 m/s east collides and sticks to a 5.0 kg toy magnetic car also traveling at 2.0 m/
Afina-wow [57]

Answer:

2.44 m/s due East

Explanation:

From the question given above, the following data were obtained:

Mass of 1st car (m₁) = 4 Kg

Velocity of 1st car (u₁) = 3 m/s

Mass of 2nd car (m₂) = 5 Kg

Velocity of 2nd car (u₂) = 2 m/s

Final velocity (v) =?

The final velocity can be obtained as follow:

v(m₁ + m₂) = m₁u₁ + m₂u₂

v(4 + 5) = (4×3) + (5×2)

9v = 12 + 10

9v = 22

Divide both side by 9

v = 22/9

v = 2.44 m/s

Thus, the final velocity is 2.44 m/s.

Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.

4 0
3 years ago
Describe the link between the distance from the Sun and the speed
tresset_1 [31]

Answer: The closer the planet is to the sun, the faster it travels

Explanation:

The speed of the planets goes down as the chart shows planets farther away from the sun

6 0
3 years ago
76.8 kPa to mm of Hg
Bumek [7]

Answer:

The answer is 576.0473

Explanation:

Hope this helps.

Please mark my answer as brainliest?

3 0
3 years ago
Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th
Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

\theta=30^{\circ}

So, the ball's horizontal velocity is

v_x = u cos \theta = (15)(cos 30)=13.0 m/s

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

y=u_yt+\frac{1}{2}at^2

where

u_y = u sin \theta = (15)(sin 30) = 7.5 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

d=v_x t =(13.0)(1.19)=15.5 m

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

d=98 m

while the horizontal velocity of the ball is

v_x = u cos \theta = (34)(cos 30)=29.4 m/s

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

y= h + u_y t + \frac{1}{2}at^2

where

h = 1.2 m is the initial height

u_y = u sin \theta = (34)(sin 30)=17.0 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

Substituting t = 3.33 s,

y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

u_y' = 7 m/s

So its position of the glove at time t' is

y'= h' + u_y' t' + \frac{1}{2}at'^2

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

t'' = t -t'

So we have two solutions:

t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he  falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
Which bright solar feature is shown in the picture above?
Ghella [55]

Answer : (B) Prominence

Explanation :

A large, glittering and gaseous characteristic which is extending outward from the surface of the sun is called <em>Prominence</em>.

<em>Photosphere</em> is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

<em>Corona</em> is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

6 0
3 years ago
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