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3 years ago

What kind of stored energy enables a stretched bow to shoot an arrow?

Physics

2 answers:

blsea [12.9K]3 years ago

7 0

Sorry if im wrong but I'll say kinetic energy

Hope it helps:)

Aliun [14]3 years ago

6 0

Stretching increases potential of the bow since its elastic.

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

2 years ago

What is the momentum of a 200 kg truck travelling at 20 m/s?

geniusboy [140]

Answer:

p = 4000 kg-m/s

Explanation:

Given that,

The mass of a truck, m = 200 kg

Speed of the truck, v = 20 m/s

We need to find the momentum of the truck. The formula for momentum is given by :

p = mv

so,

$p=200\times 20\\\\p=4000\ kg-m/s$

So, the momentum of the truck is equal to 4000 kg-m/s.

8 0

2 years ago

Me podrían dar la respuesta y el proceso?

Nat2105 [25]

Answer:

I don’t understand Espanol

Explanation:

sorry

7 0

2 years ago

A 5.00 kg mass is placed on top of a vertical spring, which compresses a distance of 3.13 cm. Calculate the force constant (in N

77julia77 [94]

Answer:

Explanation:

Step one:

given data

mass m=5kg

compression x= 3.13cm to m= 0.0313m

<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>

F=ke

where

k= spring constant in N/m

e= extension/compression in

Step two:

assume g= 9.81m/s^2

F=mg

F=5*9.81

F=49.05N

substitute in the expression F=ke

49.05=k*0.0313

k=49.05/0.0313

k=1567.09 N/m

<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>

8 0

2 years ago

When two license plates are issued they need to be attached to the _________ of the vehicle they were issued for.

Sergio039 [100]

Answer:

Front and the rear bumpers.

Explanation:

When two license plates are issued they need to be attached to the Front and the rear bumpers of the vehicle they were issued for.

7 0

3 years ago