All categories

3 years ago

A lightning bolt with 13 kA strikes an object for 14 μ s. How much charge is deposited on the object?

Physics

1 answer:

pogonyaev3 years ago

8 0

Answer:

0.182C

Explanation:

Using Q= It

= 13x10^3 . 14x10^-6

= 0.182C

You might be interested in

A pair of toy freight cars, one twice the mass of the other, fly apart when a compressed spring that joins them is released. Acc

saul85 [17]

Answer:

greater acceleration is experienced by the car with lower mass

Explanation:

Since both the toys are connected by same spring so the force due to spring on both the toys will be same and it is given as

$F = kx$

now we know by Newton's II law

$F = ma$

so here we have

$a = \frac{F}{m}$

here we have same force on both the blocks

so acceleration will be more if mass is less

so greater acceleration is experienced by the car with lower mass

8 0

3 years ago

What can parents do if they find a mistake in the student records? A. They can’t do anything B. Make a correction themselve

Lelechka [254]

D. Ask the school to make the correction

5 0

3 years ago

Read 2 more answers

Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?

Mademuasel [1]

The first opiton is the answer A)<span>Rahul’s weight
</span>

8 0

3 years ago

Read 2 more answers

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa, т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P ) = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅ , where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅ of CD on the shaft BC

Torque = 660 * 0.3 cos∅

= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0

3 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately

Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second

3 0

3 years ago