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2 years ago

1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts

Physics

1 answer:

SashulF [63]2 years ago

5 0

Answer:

by using it's buoyant or floating effect by Archimedes.

the buoyant force act on the astronauts body and make he/ she feels like in low gravity.

the buoyant force equation is

F = Density of liquid x earth gravitational field x volume of astronauts body and suit.

the Weight of astronauts in the pools will be less than in the land or air.

Weight in water = weight in air/land - buoyant force

so the astronauts will feel like in the outer space with low gravity.

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An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if

maxonik [38]

The electron is accelerated through a potential difference of $\Delta V=780 V$ , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
$\frac{1}{2}mv^2 = e \Delta V$
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
$\Delta V$ is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
$v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s$

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
$evB=m \frac{v^2}{r}$
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
$B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T$

3 0

3 years ago

A mover brings a box up the stairs in 10 seconds. If he applied a force of 20 N over a distance 10 m on the box, calculate the p

Softa [21]

Answer:

20 Watts

Explanation:

Work = force × distance

W = (20 N) (10 m)

W = 200 Joules

Power = work / time

P = 200 J / 10 s

P = 20 Watts

8 0

2 years ago

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the sys

Archy [21]

Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.

<h3>Conservation of energy</h3>

The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B$

$v_B = \frac{2v_0}{3} \ m/s \ to \ the \ left$

At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.

The complete question is below:

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?

Learn more about conservation of energy here: brainly.com/question/166559

5 0

1 year ago

Read 2 more answers

A rocket travels 1.3 km in 62 ms. What is its average speed in m⋅s−1? Do not give your answer in scientific notation. The answer

hjlf

Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

$Speed = \frac {distance}{time}$

$Speed = \frac {1300}{3720}$

Average speed = 0.35 m/s

7 0

3 years ago

The diagram shows a position-time graph What is the displacement of the object

algol13

The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.

5 0

2 years ago