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2 years ago

If you hear thunder 5 seconds after you see a lightning bolt how far is the storm

Physics

2 answers:

Ivanshal [37]2 years ago

7 0

The storm is about 1 mile away, every 5 seconds is approximately 1 mile. Hope this helps!

NikAS [45]2 years ago

4 0

I think the answer is 3 miles because its storming now where I live

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2 years ago

A student places her 490 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by

Paul [167]

Answer:

book speed is 3.99 m/s

Explanation:

given data

mass m = 490 g = 0.490 kg

compressing x = 7.10 cm = 0.0710 m

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to find out

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solution

we know energy is conserve so

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3 years ago

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2 years ago

Read 2 more answers

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

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3 years ago