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3 years ago

9

If a person pulls back a rubber band on a slingshot without letting to go of it, what type of energy will the rubber band have?

A. Potential energy B. Rotational energy C. Kinetic energy D. Translational energy

1 answer:

Travka [436]3 years ago

7 0

Answer:

Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.

Potential energy is when an object has some sort of potential eg. for motion such as in this example.

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According to Freud, adults fixated in which stage could feel constantly out of control?

yarga [219]

<span>AnalStage. </span>According to Freud, a<span>dults fixated in this area could feel constantly out of control or could need to be in control all the time. Pleasure focus is on the anus, which occurs when a child learns to control bladder and bowel movements. Fixation in this area can be caused from struggles during potty training.</span>

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3 years ago

How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for

kumpel [21]

As we know that electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

$F_g = 1.98 \times 10^{20} N$

here we can assume that both will have equal charge of magnitude "q"

now we have

$1.98 \times 10^{20} = \frac{kq^2}{r^2}$

$1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}$

$1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2$

now we have

$q = 5.7 \times 10^{13} C$

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3 years ago

The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co

djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

$\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\$

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0

3 years ago

A grating that has 3900 slits per cm produces a third- order fringe at a 28.0° angle. What wavelength of light is being used? Ex

Zolol [24]

Explanation:

Given that,

Number of slits per cm, $N=3900\ lines/cm=390000\lines/m$

The third fringe is obtained at an angle of, $\theta=28$

We need to find the wavelength of light used. The grating equation is given by :

$d\ sin\theta=n\lambda$

$d=\dfrac{1}{N}=\dfrac{1}{390000}$

$d=2.56\times 10^{-6}\ m$

$\lambda=\dfrac{d\ sin\theta}{n}$ , n = 3

$\lambda=\dfrac{2.56\times 10^{-6}\times \ sin(28)}{3}$

$\lambda=4.006\times 10^{-7}\ m$

$\lambda=400\ nm$

So, the wavelength of the light is 400 nm. Hence, this is the required solution.

4 0

3 years ago

A seismographic station receives S and P waves from an earthquake, separated in time by 17.8 s. Assume the waves have traveled o

Georgia [21]

Answer:

D = 230.2 Km

Explanation:

let distance between seismograph and focus of quake is D

From time distance formula we can calculate the time taken by the S wave

$T_1 =\frac{D}{4.5}$

From time distance formula we can calculate the time taken by the P wave

$T_2 =\frac{D}{6.90}$

It is given in equation both waves are seperated from each other by 17.8 sec

so we have

$T_1 - T_2 = 17.8$ sec

Putting both time value to get distance value

$\frac{D}{4.5} - \frac{D}{6.90} = 17.8$

D = 230.2 Km

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3 years ago