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agasfer [191]
4 years ago
7

Speed deamplifier. Block 1 of mass m1 slides along an x axis on a frictionless floor at speed 4.00 m/s. Then it undergoes a oned

imensional elastic collision with stationary block 2 of mass m2 = 2.00m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 4.00m1. (a) After all the collisions, what is the speed of block 3?

Physics
1 answer:
alisha [4.7K]4 years ago
7 0

Answer:

29J

Explanation:

Please see attachment .

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What is the relationship between the strength of an
Anastaziya [24]

Answer:

As the number of turns in the coil increases, the strength  of the electromagnet increases.

Explanation:

When current flows through a coil the coil behaves as an electromagnet. The strength of electromagnet depend the amount of current, no of turns of coil and the core of coil.

B=μ₀ N I

μ₀ = permeability of the core

N = Number of turns of the coil

I = Current flowing through the coil

Increasing the current and number of coils increase the strength of electromagnet.

3 0
4 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
4 years ago
Which is the best example of the concept of heredity? A. A black cat blends into its environment at night. B. A baby has his mot
Hitman42 [59]
B) A baby has his mother's hair and eye color 
4 0
3 years ago
Read 2 more answers
A new band sensation is playing a concert and recording it for a live album to be released this summer. The band asks the sound
Amanda [17]

The sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

<h3>Amplitude of sound wave</h3>

The amplitude of a sound wave is the maximum vertical displacement of the sound wave.

The sound mixer will need to increase the amplitude of the sound wave produced by the singer.

The increase in the amplitude of the sound wave produced by the lower tune singer will result in increased loudness of the sound.

Thus, the sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

Learn more about sound waves here: brainly.com/question/1199084

8 0
2 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
4 years ago
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