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PIT_PIT [208]
3 years ago
5

During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros

econds . How much charge is transferred from the cloud to the earth during such a strike?
Physics
1 answer:
dangina [55]3 years ago
6 0

Answer:

1 C

Explanation:

The intensity of electric current is defined as

I=\frac{q}{t}

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

I=2.50\cdot 10^4 A is the current

t=40.0 \mu s = 40.0\cdot 10^{-6} s is the time interval

Solving the formula for q, we find the amount of charge transferred:

q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C

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A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy
m_a_m_a [10]

Answer:

(A) Spring constant will be 126.58 N/m

(B) Amplitude will be equal to 0.177 m

Explanation:

We have given mass of the block m = 200 gram = 0.2 kg

Time period T = 0.250 sec

Total energy is given TE = 2 J

(A) For mass spring system time period is equal to T=2\pi \sqrt{\frac{m}{K}}

So 0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}

0.0398=\sqrt{\frac{0.2}{K}}

Now squaring both side

0.00158=\frac{0.2}{K}

K = 126.58 N/m

So the spring constant of the spring will be 126.58 N/m

(B) Total energy is equal to TE=\frac{1}{2}KA^2, here K is spring constant and A is amplitude

So 2=\frac{1}{2}\times 126.58\times A^2

A^2=0.0316

A = 0.177 m

So the amplitude of the wave will be equal to 0.177 m

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3 years ago
Which of these forces are acting within a nucleus? Check all that apply.
zavuch27 [327]

Answer:

Excuse me what?

Explanation:

5 0
3 years ago
What charges form a neutral charge?
brilliants [131]

A negative to a negative charge will make a neutral charge.

4 0
3 years ago
Read 2 more answers
Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal
Harrizon [31]

Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

8 0
3 years ago
Neglecting air resistance, what maximum height will be reached by an arrow launched straight upward with an initial speed of 35
tankabanditka [31]
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.

The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)

Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
3 0
3 years ago
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