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3 years ago

In which direction would the circle go if it was in between a positive and negative charge

Physics

1 answer:

evablogger [386]3 years ago

7 0

It would go to the positive charge im like 99% sure of that

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A car sitting at a red light begins to accelerate at 2.0 m/s2 when the light turns green. It continues with this acceleration un

KonstantinChe [14]

Answer:

$d = 820 m$

Explanation:

here we know that car starts from rest and continue to accelerate till it will reach to maximum speed of 20 m/s

so we will have

$v_f - v_i = at$

$20 - 0 = 2t$

$t = 10 s$

so car will accelerate till t = 10 then it will move with uniform speed

so the distance moved by the car till it accelerates is given as

$d = \frac{1}{2}at^2$

$d = \frac{1}{2}(2)(10^2)$

$d_1 = 100 m$

now it will move with uniform speed for next 36 s

so we have

$d_2 = 20(36) = 720 m$

so total distance moved by the car is given as

$d = d_1 + d_2$

$d = 100 + 720$

$d = 820 m$

8 0

3 years ago

What is the name of the process that is responsible for the formation of elements heavier than hydrogen? A) fusion B) fission C)

gladu [14]

Answer:

A) Fusion

Explanation:

Fusions of atoms heavier than hydrogen like helium takes place in the core of the star, at the core the temperature and density of gases is the greatest .

5 0

3 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

3 years ago

the train is travelling at 15 m/s. It accelerates at 3 m/s² for 20 seconds. How far does the train travel in this time?

Anvisha [2.4K]

18meters per 20 seconds so

5 0

3 years ago

Read 2 more answers

What are the components of the "Earth Radiation Budget?"

Alex777 [14]

Answer:

The energy entering, reflecting, absorbed, and emitted by the earth system are the components of the Earth's radiation budget.

Explanation:

I hope this helps also I hope you have a great day and a new year.

4 0

3 years ago