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Helen [10]
2 years ago
5

The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u

pthrust force.
The hot-air balloon descends vertically at constant speed. The force of air resistance on the balloon is F.
Which weight of material must be released from the balloon so that it ascends vertically at the same constant speed?
Physics
1 answer:
Aleksandr [31]2 years ago
3 0
Let
upthrust = T
weight = W = mg
Air resistance = F

When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)

Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)

Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
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A simple generator has a square armature 6.0 cm on a side. The armature has 85 turns of 0.59-mm-diameter copper wire and rotates
FrozenT [24]

Answer:

f=15.5 Hz

Explanation:

Let's determine the internal resistance:

R=\frac{(p*L)}{A}

ρ = 1.68*10^-8 Ω m

L=0.060m*4*60 = 14.4m

A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2

R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884Ω

Since the bulb is rated at 12.0 V and 25.0 W,

Current

I=\frac{25W}{12.0v}=2.08 A

Therefore, voltage drop inside generator =

V=(2.08 A)*(0.88)=2.35v

Actual EMF required is

E_{mf}=12.0v+2.35v=14.35v

Note that this is an RMS value.  

The peak voltage is

v_{peak}=14.15v*\sqrt{2} =20.29v

For a generator, by Faraday's Law,

E_{(max)}=N*B*A*w

20.29v=(60)*(0.650T)*(0.06m)^2*ω

ω=144.5\frac{rad}{s}

f=ω/(2π)=

f=144.5 rad/s/(2π)

f=23.001 Hz

6 0
2 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
This table shows Wayne’s weight on four different planets. Planet Wayne’s weight (pounds) Mars 53 Neptune 159 Venus 128 Jupiter
oee [108]
The correct order is (in decreasing order of gravity strength)
Jupiter - Neptune - Venus - Mars

In fact, Wayne's weight on each planet is given by
W=mg
where m is Wayne's mass, which is a constant value, and g is the gravity strength at the surface of the planet. Therefore, the Wayne's weight W on each planet is directly proportional to the gravity strength of that planet: so the planet with the strongest gravity is the one where Wayne's weight is the greatest (Jupiter, 333 pounds), followed by Neptune (159),  Venus (128) and Mars (53).
8 0
2 years ago
A boy jumps into an indoor swimming pool. He notices that the water appears to get colder as he
Viefleur [7K]

Answer:

cold air is more dense than warm water so it sinks to the bottom of the pool

7 0
1 year ago
If a 110 kg go-cart traveling at a velocity of 13.41 m/s has a collision with an impulse of 615 Nxs, what is the
mafiozo [28]

Answer:

5.59 m/s

Explanation:

We are given;

Mass = 110 kg

Initial velocity: u = 13.41 m/s

Force = 615 N

Time(t) = 1 s

Now, the formula for force is;

Force = mass x acceleration

Thus;

615 = 110 × acceleration

\Acceleration(a) = 615/110 = 5.591 m/s²

Now, using Newton's first law of motion, we can find acceleration (a). Thus;

v = u + at

v = 13.41 + (5.591 × 1)

v ≈ 19 m/s

So,the change in velocity is;

Final velocity(v) - Initial velocity(u) = 19 - 13.41 = 5.59 m/s

6 0
2 years ago
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