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Step2247 [10]
3 years ago
15

Identify the situations that have an unbalanced force. Check all that apply. A baseball speeds up as it falls through the air. A

soccer ball is at rest on the ground. An ice skater glides in a straight line at a constant speed. A bumper car hit by another car moves off at an angle. A balloon flies across the room when the air is released.
Physics
2 answers:
viktelen [127]3 years ago
5 0
<span>A baseball speeds up as it falls through the air.
Yes.  Forces on the balloon are unbalanced. 
The balloon is speeding up, so we know that the downward force
of gravity is stronger than the upward force of air resistance.

A soccer ball is at rest on the ground.
No. The ball is not accelerating, so we know that the forces on it
are balanced.
The downward force of gravity on the ball and the upward force
of the ground are equal.


An ice skater glides in a straight line at a constant speed.
No. The skater's speed and direction are not changing, so he is not
accelerating.  That tells us that the forces on him are balanced.

A bumper car hit by another car moves off at an angle.
Yes.  The direction in which the car was moving changed. 
That's acceleration, so we know that the forces on it are unbalanced,
at least at the moment of impact. 

A balloon flies across the room when the air is released.
Yes.  The balloon was not moving. But when the little nozzle was
opened, it started to zip around the room.  So its speed changed.
And, as it goes bloozing around the room, its direction keeps changing too. 
There's a whole lot of acceleration going on, so we know the forces on it
are unbalanced.</span>
Norma-Jean [14]3 years ago
5 0
A bumper car hit by another car moves off at an angle.<span>A balloon flies across the room when the air is released.
</span><span>A baseball speeds up as it falls through the air.
</span>
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by conduction, which is direct touch

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A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int
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Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of

<em>F</em> = <em>m a</em>   →   <em>a</em> = <em>F </em>/ <em>m</em>

so that the cart's final speed is

<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>

<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>

<em />

If we force is halved, so is the accleration:

<em>a</em> = <em>F</em> / <em>m</em>   →   <em>a</em>/2 = <em>F</em> / (2<em>m</em>)

So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give

(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>

3 0
2 years ago
Aluminum allows for the flow of electrons, while glass does not. Would an aluminum wire surrounded by glass be an effective desi
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no

Explanation:

I do not think that I would because even though its a conductor in the insulator I think it would insulate it before it will work (not sure if that makes sense)

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Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density o
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Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

8 0
3 years ago
5) Consider pushing a 50.0 kg box through a 5.00 m displacement on both a flat surface and up a
Svetach [21]

a) The work done by the gravitational force on the flat surface is zero

b) The work done by the gravitational force on the ramp is -634 J

c) The work done by the applied force on the flat surface is 500 J

d) The work done by the applied force on up along the ramp is 500 J

Explanation:

a)

The work done by a force is given by the equation

W=Fdcos \theta

where

F is the magnitude of the force

d is the dispalcement of the object

\theta is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

We immediately notice that the gravitational force acts downward, while the displacement is horizontal: therefore, the angle between force and displacement is 90^{\circ}; this means that cos 90^{\circ}=0, and therefore, the work done is zero:

W=0

b)

In this case, the box is pushed along the ramp. We have:

F=mg=(50.0)(9.8)=490 N is the magnitude of the force of gravity, where

m = 50.0 kg is the mass of the box

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

d = 5.00 m is the displacement of the box along the ramp

The ramp is inclined to the horizontal by 15.0^{\circ}, therefore the angle between the force of gravity and the displacement of the box (moving up along the ramp) is:

\theta=90^{\circ}+15^{\circ}=105^{\circ}

Therefore, the work done by gravity in this case is:

W=(490)(5.00)(cos 105^{\circ})=-634 J

c)

In this case, we want to calculate the work done by the force you apply as the box is pushed across the flat ground.

Here we have:

F = 100.0 N (force applied)

d = 5.00 m (displacement of the box)

\theta=0^{\circ} (the force is applied parallel to the flat surface, therefore force and displacement have same direction)

Therefore, the work done by the force you apply on the flat ground is:

W=(100.0)(5.00)(cos 0^{\circ})=500 J

d)

In this last case, we want to calculate the work done by the force you apply as the box is pushed up along the ramp.

This time we have:

F = 100.0 N (force applied is the same)

d = 5.00 m (displacement of the box is also the same)

\theta=0^{\circ} (the force is applied parallel to the ramp, therefore force and displacement have again same direction)

Therefore, the work done by the force you apply while pushing the box along the ramp is:

W=(100.0)(5.00)(cos 0^{\circ})=500 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
3 years ago
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