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2 years ago

6

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 10.0 cm away. The force of attraction is

then 20.0 mN. Find magnitude of Q, give answer in C.

1 answer:

victus00 [196]2 years ago

5 0

Answer:

a charge Q is transferred from an initially uncharged

Explanation:

Hope this helps!

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Which type of wave results from a moderately sloping coastal region?

Naddik [55]

Hey there!

Your answer: Spilling breaker

Spilling breaker usually occurs when a beach or ocean is flat, and as the waves of the wind continues to happen, slowly the region would eventually become a slope.

It's almost like play-dough. Let's say that we set a perfect flat surface of play-dough on the table. As we continue slide our hands one direction, doesn't the play dough have more on one side than the other? It eventually contains a slope if you add enough from the first place.

Your answer: Spilling breaker

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3 years ago

Read 2 more answers

Give three factors which are responsible for the vanishing forest

slava [35]

Answer:

1. Huge wildfires

2. Deforestation

3. Reduced amount of aforestation, etc

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3 years ago

An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti

Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
So, we can write the following expression for Fg:

$F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)$

b)

The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

$F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N (2)$

c)

According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

$a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2 (3)$

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3 years ago

I don’t understand I need help asap

hammer [34]

For the first question, you got them right, for the two you left blank, initial(beginning) velocity: 2 m/s the final velocity is: 12 m/s

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3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

3 years ago