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4 years ago

A pulling force is called what? A. normal B. tension C. balanced D. compression

Physics

2 answers:

worty [1.4K]4 years ago

5 0

What i think the pulling forced called is , balanced force.

joja [24]4 years ago

3 0

There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)

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An artificial satellite in a low orbit circles the earth every 98 minutes. what is its angular speed in rad/s?

postnew [5]

To finish one orbit it will take 98 x 60 seconds. So; (2 x pi)/(98 x 60) = 1.07 x 10^-3 rad/sec.

8 0

4 years ago

You are to drive to an interview in another town, at a distance of 310 km on an expressway. The interview is at 11:15 a.m. You p

Sloan [31]

Answer:

Explanation:

Time to cover first 100 km = 1 hour.

time remaining = 3.15 - 1 = 2.15 hour .

Time to cover next 42 km = 1 hour .

Time remaining = 2.15-1 = 1.15 hour.

Distance to be covered = 310 - 142

= 168 km

least speed needed = distance remaining / time remaining

= 168 / 1.15

= 146.08 km / h .

4 0

4 years ago

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

3 years ago

Bob weighs 80 pounds. He ran up the steps which rose 14 feet. Fred weighs 110 pounds. He ran up the same steps as Bob. If both b

pav-90 [236]

Bob gained (80lbs x 14ft) = 1120 ft-lbs of energy.

Fred gained (110lbs x 14ft) = 1540 ft-lbs of energy

Since they both took the same amount of time, Fred's power (rate
of doing work) was greater than Bob's power (rate of doing work).

6 0

3 years ago

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Which of the following is not an example of projectile motion?

Ilia_Sergeevich [38]

What are the answers?
in order to get help i would need the answers :)

4 0

4 years ago

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