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Gnom [1K]
3 years ago
8

What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward

-directed electric field of magnitude 670 N/CN/C ?
Physics
1 answer:
Yuri [45]3 years ago
6 0

Answer:

q = 2.067 \times 10^{-5}\ C

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

q =\dfrac{mg}{E}

q =\dfrac{0.00141 \times 9.81}{670}

q = 2.067 \times 10^{-5}\ C

Charge of the particle is equal to q = 2.067 \times 10^{-5}\ C

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A particle moves along the x axis. It is intially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.32
Nataliya [291]

Answer:

The position of the particle is -2.34 m.

Explanation:

Hi there!

The equation of position of a particle moving in a straight line with constant acceleration is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the particle at a time t:

x0 = initial position.

v0 = initial velocity.

t = time

a = acceleration

We have the following information:

x0 = 0.270 m

v0 = 0.140 m/s

a = -0.320 m/s²

t = 4.50 s  (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).

Then, we have all the needed data to calculate the position of the particle:

x = x0 + v0 · t + 1/2 · a · t²

x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²

x = -2.34 m

The position of the particle is -2.34 m.

6 0
3 years ago
About how much of Earth's water is available for humans to use?
lesya692 [45]

Answer:

0.3 %

Explanation:

Earth cleans and replenishes the water supply through the hydrologic cycle. The earth has an abundance of water, but unfortunately, only a small percentage, is even usable by people.

5 0
3 years ago
Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v
bazaltina [42]

Answer:

1. t_2 = 2t_1

2. t_2 = t_1\sqrt{2}

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

a_1 = 2a_2

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

t = v / a_1

Now that acceleration is halved:

t = \frac{v}{2a_2}

\frac{v}{a_2} = 2t

You would need to push for twice amount of time t_2 = 2t_1

2. The distance traveled by the puck is as the following equation:

d = at^2

So if the acceleration is halved while maintaining the same d:

\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}

As d_1 = d_2, then d_1/d_2 = 1. Also a_1 = 2a_2

1 = \frac{2a_2t_1^2}{a_2t_2^2}

t_2^2 = 2t_1^2

t_2 = t_1\sqrt{2}\approx 1.14t_1

So t increased by 1.14

7 0
3 years ago
Any one tell me about the earth rotation it rotatining or not with any proof? ​
Dafna11 [192]
The proof that the earth is rotating is the happens of night and day also the seasons, eg. winter, summer, autumn.
6 0
3 years ago
a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr
Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}} where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz

5 0
3 years ago
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