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3 years ago

15

A force is applied to a 2.9 kg mass and pro-

1 answer:

adoni [48]3 years ago

3 0

Answer:

5.8 or 5.77

Explanation:

I'm not sure

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A speeding motorist traveling 120 km/h passes a stationary police officer. The officer immediately begins pursuit at a constant

Lelu [443]

120 km/h = 33.33 m/s 10.8 km/h/s = 3 m/s/s the motorist will cover a distance of 33.33 m x T(seconds) the police officer will cover a distance of 1/2 (3) T^2 they will be at the same point when 33.33T = 1.5 T^2 33.33 = 1.5 T 3T = 66.66 T = 22.22 seconds it will take the officer 22.22 seconds to catch the motorist. the officer will be moving V=AT = 3m/s x 22.22 seconds = 66.66 m/s almost 240 km/h (239.976 km/h)

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4 years ago

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Vesnalui [34]

Answer:

Observation 1 represents a solar wind because solar winds are a continuous stream of particles given off by the sun.

Explanation:

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3 years ago

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Solenoid 2 has twice the diameter, twice the length, and twice as many turns as solenoid 1. How does the field B2 at the center

klemol [59]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is option 3

Explanation:

From the question we are told that

The diameter of solenoid 1 is $d_1$

The length of solenoid 1 is $L_1$

The number of turns of solenoid is $N_1$

The diameter of solenoid 2 is $d_2 = 2d_1$

The length of solenoid 2 is $L_2 = 2L_1$

The number of turns of solenoid 2 is $N_2 = 2 N_1$

Generally the magnetic in a solenoid is mathematically represented as

$B = \frac{\mu_o * N * I }{L}$

From this equation we see that

$B \ \alpha \ \frac{N}{L}$

$B = C \frac{N}{L}$

Here C stands for constant

=> $C = \frac{B * \frac{L}{N}$

=> $\frac{B_1 * \frac{L_1}{N_1} = \frac{B_2 * \frac{L_2}{N_2}$

=> $\frac{B_1}{B_2 } = \frac{N_1 L _2}{ N_2L_1}$

=> $\frac{B_1}{B_2 } = \frac{N_1 * (2 L_1)}{ (2 N_2)L_1}$

=> $\frac{B_1}{B_2 } = 1$

=> $B_2 = B_1$

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4 years ago

2. Our solar system is made up of the Sun, 8 planets, and other bodies such as

Pie

It’s 25 because I already took the testing

8 0

3 years ago

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

Vlad [161]

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

4 0

3 years ago