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2 years ago

An observer notices that 12 waves pass by in 4 seconds. What is the frequency of these waves? 0.33 Hz 16 Hz 48 Hz 3 Hz

Physics

1 answer:

Dennis_Churaev [7]2 years ago

8 0

Answer:

Frequency = 3 Hz

Explanation:

Frequency is a measure of Hertz. Recall that Hertz is the unit expressing cycles/second, where one second is the denominator of the fraction for simplicity. If there are 12 waves every four seconds, and one wave represents one cycle, dividing 12 waves by 4 seconds will give the answer of 3 waves (or cycles) per one second.

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When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

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

here we know that

$q_1 = +8.4\mu C$

$q_2 = +5.6 \mu C$

force between two charges is given as

$F = 0.66 N$

now we have

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

$0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}$

$r = 0.8 m$

so it is separated by 80 cm distance

3 0

2 years ago

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An ant does 1 N-m of work in dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar.

algol13

The ant would drag the sugar 500m

Good luck!

8 0

2 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

2 years ago

What are the parts of an atom

Ivenika [448]

Number three

They contain protons (positive), neutrons (negative), electrons (neutral) and all are in a nucleus which is part of an atom

7 0

2 years ago

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The sun emits electromagnetic waves with a power of 4.0 × 10²⁶ W. Determine the intensity of electromagnetic waves from the sun

Sphinxa [80]

Answer:

I_v = 2,700 W / m^2

I_m = 610 W / m^2

I_s = 16 W / m^2

Explanation:

Given:

- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W

- Radius of Venus r_v = 1.08 * 10^11 m

- Radius of Mars r_m = 2.28 * 10^11 m

- Radius of Saturn r_s = 1.43 * 10^12 m

Find:

Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.

Solution:

- We know that Power is related to intensity and surface area of an object follows:

I = P / 4*pi*r^2

Where, A is the surface area of a sphere models the atmosphere around the planets.

- The intensity at the surface of Venus is calculated as:

I_v = P_s / 4*pi*r^2_v

I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2

I_v = 2,700 W / m^2

- The intensity at the surface of Mars is calculated as:

I_m = P_s / 4*pi*r^2_m

I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2

I_m = 610 W / m^2

- The intensity at the surface of Saturn is calculated as:

I_s = P_s / 4*pi*r^2_s

I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2

I_s = 16 W / m^2

7 0

3 years ago