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kramer
3 years ago
9

PLEASE HELP WILL MARK BRAINLIEST

Physics
2 answers:
valentina_108 [34]3 years ago
6 0

Answer: The charges will become polarized with a positive zone near the balloon

Explanation: I just took this quiz

tester [92]3 years ago
4 0

Answer:

A The overall charge will become negative as positive charges jump to the balloon

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Which type of storm is measured by the fujita scale?
Serggg [28]
It's used for the intensity of a tornado
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3 years ago
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A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total
Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

h_2 = \frac{4}{5}h

so the distance covered is given as

d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........

here we know that

h = 10 feet

d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)

d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})

d = 90 ft

8 0
3 years ago
A moving walkway at an airport has a speed v1 and a length L. A woman stands on the walkway as it moves from one end to the othe
bearhunter [10]

Answer:

t=L/V_1

Explanation:

<u>solution:</u>

Let E be an observer, and B a second observer traveling with velocity V_{BE}  as measured by E. If E measures the velocity of an object A as V_{AE}  then B will measure A velocity as

V_{AB} = V_{AE} -  V_{BE}  

Applied here,

the walkway (W) and the man (M) are moving relative to Earth (E}, the velocity of the man relative to the moving walkway is

V_{MW} = V_{ME} -  V_{WE}  

V_1=V_{WE},

V_2=V_{MW}

The time required for the woman, traveling at constant speed V_1 relative to the ground, to travel distance L relative to the ground is :

t=L/V_1

 

3 0
4 years ago
A 68 kg runner exerts a force of 59N. What is the acceleration of the runner?
Korvikt [17]

Explanation:

<em><u>Newton's First Law ,</u></em>

F = ma

a = m/F

a = 68 / 59

a = 1.15 m/s2

4 0
4 years ago
Josh starts his sled at the top of a 2.9-m-high hill that has a constant slope of 25∘. After reaching the bottom, he slides acro
algol13

Answer:

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2(9.81)(2.9)

   =56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v²- u² = 2 -μ g S

v=0 m/s

-(7.54)²=-2(0.06)(9.81)S

S=48.29 m

3 0
3 years ago
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