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3 years ago

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

Physics

1 answer:

SashulF [63]3 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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You work at a retail store. Recently, paychecks have had errors and there have been many employee complaints. You have been assi

Nata [24]

Answer:

C,B,A

Explanation:

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3 years ago

Read 2 more answers

An automobile with 0.500 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, negle

Zinaida [17]

Answer:

n = 25,464,790.89 revolutions

The tires have made 25,464,790.89 revolutions

Explanation:

Given;

Radius of tires r = 0.5 m

Total distance travelled d = 80,000 km = 80,000,000 m

1 revolution = 2πr

Total distance d = number of revolutions n × 2πr

d = n×2πr

d = 2πnr

Making n the subject of formula;

n = d/2πr

Substituting the given values;

n = (80,000,000)/(2×π×0.5)

n = 25,464,790.89470 revolutions

n = 25,464,790.89 revolutions

The tires have made 25,464,790.89 revolutions

5 0

4 years ago

calculate the electric field strength between two parallel plates separated by 0.50 cm, across which is a potential of 12 volts.

Alexxandr [17]

Formula for Electric Field strength , E = V/d

where V = Voltage in volts, and d = distance of separation in meters.

d = 0.5 cm = 0.005 m, V = 12 V

E = V/d = 12 / 0.005

E = 2400

Electric Field Strength = 2400 Volts/meter

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3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago

Name 5 ways you can deal with a bully without engaging in a physical altercation with him/her.

Delicious77 [7]

Answer:

ignore the bully ,tell the bully to stop,make a joke or laugh with the bully,stick with friend,know how to get out of the bullying situation

Explanation:

6 0

3 years ago