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dalvyx [7]
3 years ago
15

Sketch a free body diagram for a hot air balloon which is rising off the ground. Indicate the magnitude of the force by utilizin

g a scale of length for the force vectors. Include your scale and Describe one Action-Reaction Pair (in full sentence form below your image)
Physics
1 answer:
vaieri [72.5K]3 years ago
3 0

Answer:

   F_net = B -W

The forces of action and reaction are the weight of the balloon that is the force of attraction of the Earth and the outside so the balloon pulls the earth that has an upward direction and is applied to the planet

Explanation:

A hot air balloon is subjected to the force of its weight directed towards the center of the Earth. The thrust due to the cold air released, this thrust is directed upwards.

If we assume that the balloon rises at a constant speed

         F_net = B -W

         F_neta = ρ g V_body - ρ_body g V_body

The forces of action and reaction are the weight of the balloon that is the force of attraction of the Earth and the outside so the balloon pulls the earth that has an upward direction and is applied to the planet

You might be interested in
A tennis ball is dropped from 1.13 m above the
Alex73 [517]

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

7 0
3 years ago
An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of
Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

5 0
2 years ago
21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest
zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

mgh=\frac{1}{2}mu^2

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

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

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}

where \Delta p is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

\Delta t=10.0 ms =0.010 s is the duration of the collision

Substituting,

F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N

And the negative sign means the direction of the force is opposite to the motion (so, upward).

6 0
3 years ago
two circular loops of wire, each containing a single turn, have the same radius of 4.0 cm and a common center. the planes of the
REY [17]

The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.

We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as

B = μ₀ . I / 2r

where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.

From the question above, we know that:

r = 4 cm = 0.04 m

I = 1.7 A

By substituting the parameter, we get

B = μ₀ . I / 2r

B = 4π×10¯⁷ . 1.7 / (2.0.04)

B = 2.67 x 10¯⁵ T

Due to the perpendicular plane of loops, the total magnetic field at center will be

Btotal = √(2(B²))

Btotal = √(2(2.67 x 10¯⁵²))

Btotal = 3.78 x 10¯⁵ T

Find more on magnetic field at: brainly.com/question/7802337

#SPJ4

3 0
2 years ago
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