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3 years ago

HEY CAN ANYONE PLS ANSWER THESE QUESTION!!!!!

Physics

1 answer:

Pie3 years ago

5 0

1. A model can come in many shapes, sizes, and styles.

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2. Models can help scientists communicate their ideas, understand processes, and make predictions.

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3. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons.

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4. The plum pudding model has electrons surrounded by a volume of positive charge

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5. Electrons are kept in the orbit around the nucleus by the electromagnetic force.

IamSugarbee

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An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit

nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

5 0

3 years ago

A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows:

$F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB$ (1)

Where:

m: is the proton's mass = 1.67*10⁻²⁷ kg

v: is the proton's velocity

r: is the radius of the proton's orbit

q: is the proton charge = 1.6*10⁻¹⁹ C

B: is the magnetic field = 0.040 T

Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$ (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$ (3)

Where:

K: is kinetic energy

U: is electrostatic potential energy

ΔV: is the potential difference = 1.0 kV

Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!

5 0

3 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

3 years ago

I need help with questions b and d, that’s all. Thank you.

Mkey [24]

b). The power depends on the RATE at which work is done.

Power = (Work or Energy) / (time)

So to calculate it, you have to know how much work is done AND how much time that takes.

In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.

The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.

Without knowing the time, we can't calculate the power.

...

d). Kinetic energy = (1/2) · (mass) · (speed squared)

On the way up, the car stops when it reaches point-A.

On the way down, the car leaves point-A from "rest".

WHILE it's at point-A, it has no speed. So it has no (zero) kinetic energy.

7 0

2 years ago

What are the different skills and strategies were you aware of in this situation that indicated to you that it would be appropri

const2013 [10]

Answer:

dad

Explanation:

daddy

4 0

2 years ago