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cupoosta [38]
3 years ago
9

An electric field around the rubbed balloon exerts a force that pulls the paper. Identify the contact forces and the field force

s shown in the photo of the balloon

Physics
2 answers:
sleet_krkn [62]3 years ago
7 0
Contact force:Friction
Field force:Electric force
melamori03 [73]3 years ago
4 0

Answer: the contact force is when you rub the balloon on a surface

Explanation:

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Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a
Viktor [21]

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, \theta=25^{\circ}

Time, t = 3 s

Let h is the height of the hill from the horizontal,

h=l\ sin\theta

h=12\times \ sin(25)

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

P=\dfrac{E}{t}

P=\dfrac{mgh}{t}

P=\dfrac{55.8\times 9.8\times 5.07}{3}

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

3 0
3 years ago
A ball is moving in a certain direction. What could happen to the ball if a greater force was applied on the ball along its dire
alukav5142 [94]

Answer:

the one going faster would prolly stop and the one it hit would start rolling the opposite direction it was. like think about if u were playing pool.

Explanation:

8 0
2 years ago
Siobhan wants to measure the mass of a bag of flour. What should she do?
I am Lyosha [343]
<span>Hey there!
Awesome question=)

Siobhan can place it on a regular scale(0 gravity area), or she can use the "balance scale"
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7 0
4 years ago
Read 2 more answers
5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)
jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

3 0
3 years ago
While a block slides forward 1.35 m, a force pulls back at a 135 direction, doing -17.8 J of work. what is the magnitude of the
Verdich [7]

The magnitude of the force is 18.6 N

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

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

For the block in this problem, we have

W=-17.8 J is the work done

d = 1.35 m is the displacement of the block

\theta=135^{\circ} is the angle between the force and the displacement

Solving for F, we find the magnitude of the force:

F=\frac{W}{d cos \theta}=\frac{-17.8}{(1.35)(cos 135)}=18.6 N

Learn more about work here:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
4 years ago
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