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Jobisdone [24]

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

4 0

3 years ago

If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration

Nataly_w [17]

Answer:

When we double the angular velocity the maximum acceleration $(a_{max})$ will changes by a factor of 4.

Explanation:

Given the angular frequency $(\omega)$ of the simple harmonic oscillator is doubled.

We need to find the change in the maximum acceleration of the oscillator.

$a_{max}=A\omega^2$

Now, according to the problem, the angular frequency $(\omega)$ got doubled.

Let us plug $\omega=2\times \omega$ . Then the maximum acceleration will be $a_{max'}$

$a_{max}=A\omega^2$

$a_{max'}=A(2\times \omega)^2\\a_{max'}=A\times 4\omega\\a_{max'}=4A\omega$

$a_{max'}=4a_{max}$

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

6 0

3 years ago

A submarine emerges 1/9 of its volume when it partially floats on the sea surface. For make it completely submerge it is necessa

AleksAgata [21]

Answer:

4,524,660 N

Explanation:

Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.

m/9 = (1026 kg/m³) (50 m³)

m = 461,700 kg

mg = 4,524,660 N

3 0

3 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

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

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

3 years ago

Allison is looking through a telescope at an object in space. The object looks like a very small planet, and it does not have a

Verizon [17]

Answer:

Allison is probably looking at the asteroid.

Explanation:

Asteroids are the giant gas balls of inner solar system that are present in the space.
Asteroids orbits around the sun in very irregular ans strange path unlike the other planets.
This is also the reason why it's called as the minor planets.
It could have been comet but if it had been a comets he must have seen the tail but asteroids do not have any tail like structure.

8 0

3 years ago