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Georgia [21]
3 years ago
6

What does the front lens in a telescope do?

Physics
2 answers:
Daniel [21]3 years ago
7 0

Answer: Bend the light.

Explanation: ...

bixtya [17]3 years ago
6 0
They focus the light and make distant objects appear brighter, clearer and magnified.
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A racing car can complete a 900 m long lap in 15 s. What is the speed of the car?
Gemiola [76]

60 is the speed of the car

4 0
4 years ago
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2 strings both vibrate at exactly 220 Hz. The tension in one of them is then decreased sightly. As a result, 3 beats per second
MAVERICK [17]

Explanation:

Given that,

2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.

The tension in one of them is then decreased sightly, then f_2 will decrese.

Beat frequency, f=f_1-f_2

3=220-f_2

f_2=217\ Hz

So, the new frequency of the string is 217 Hz. Hence, this is the required solution.

3 0
3 years ago
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
3 years ago
Infer why the doppler effect can or can not occur in all waves.
Darya [45]

Answer:

it can occur in all waves because all wave have a frequency

8 0
3 years ago
Which statements describe velocity and acceleration? Check all that apply.
Tom [10]

Velocity is defined by rate of change in the position

which we can also write as

v = \frac{ds}{dt}

while acceleration is defined as rate of change in velocity

a = \frac{dv}{dt}

so acceleration and velocity both are rate of change in position and rate of change in velocity with respect to time respectively

out of all above statement the correct answer must be

<u>Acceleration equals change in velocity divided by time. </u>

4 0
3 years ago
Read 2 more answers
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