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Murljashka [212]
3 years ago
13

Lightning a. occurs when a positively charged cloud base induces a negative charge on the Earth's surface. b. travels most often

from cloud to ground and cannot travel from cloud to cloud. c. may be accompanied by the sound of explosively expanding hot air, called thunder. d. involves a conductive path or leader of positive ions from cloud to ground.
Physics
1 answer:
mart [117]3 years ago
8 0

Answer:

c. may be accompanied by the sound of explosively expanding hot air, called thunder.

Explanation:

Lightning is a discharge which is due to the reaction between oppositely charged charges in the clouds, or between clouds base and the Earth surface.

The motion of the cloud causes charging of clouds by friction, thus the reaction between opposite charges (jumping of charges from one cloud to another) in the cloud can lead to lightning. Also, oftentimes the bottom of a cloud is negatively charged so that this is attracted to the positive charge on the earth surface. Thus leading to a discharge called lightning.

Thus in the given question, the appropriate option is C. This implies that, lightning may be accompanied by the sound of explosively expanding hot air, called thunder.

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The box shown on the rough ramp above is sliding up the ramp.calculate the force of friction on the box
Anna [14]

We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:

F_f=\mu N

Where.

\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:

In the diagram we have:

\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}

Adding the forces in the y-direction we get:

\Sigma F_y=N-mg_y

Since there is no movement in the y-direction the sum of forces must be equal to zero:

N-mg_y=0

Now we solve for the normal force:

N=mg_y

To determine the y-component of the weight we will use the trigonometric function cosine:

\cos 40=\frac{mg_y}{mg}

Now we multiply both sides by "mg":

mg\cos 40=mg_y

Now we substitute this value in the expression for the normal force:

N=mg\cos 40

Now we substitute this in the expression for the friction force:

F_f=\mu mg\cos 40

Now we substitute the given values:

F_f=(0.2)(10\operatorname{kg})(9.8\frac{m}{s^2})\cos 40

Solving the operations:

F_f=15.01N

Therefore, the force of friction is 15.01 Newtons.

3 0
1 year ago
Forces that are equal in size but opposite in direction are called what ?
enot [183]

Answer:

I think they are called balanced forces

Explanation:

3 0
3 years ago
Read 2 more answers
A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the s
Dmitriy789 [7]

Answer:

(A) V = 9.89m/s

(B) U = -2.50m/s

(C) ΔK.E = –377047J

(D) ΔK.E = –257750J

Explanation:

The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.

This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.

See attachment below for full solution.

5 0
3 years ago
A young kid of mass m = 36 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 3.5 m. The boy
lara31 [8.8K]

Answer:

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point mg-T=\frac{mv^2}{r}

where v=velocity needed to complete circular path

Th-resold velocity is given by mg-0=\frac{mv^2}{r}

v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s

So apparent weight of boy will be zero at top when it travels with a velocity of v=\sqrt{gr}

To get the velocity at bottom conserve energy at Top and bottom

At top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at Bottom E_b=\frac{mv_0^2}{2}

Comparing two as energy is conserved

v_0^2=4gl+gl

v_0^2=5gL

v_0=\sqrt{5gL}=13.09 m/s

Apparent weight at bottom is given by

W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N

6 0
3 years ago
How much is a city of a body when it covers 600m in 5 min?
Fofino [41]

Answer:

2m/s

Explanation:

5min × 60sec

=300

now,

600÷300

=2

3 0
3 years ago
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