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adoni [48]
2 years ago
10

A bored college student on top of a 6-story tall building drops a water balloon on his friends directly below. In one second it

falls one story down from the top. In one more second it will be:
a Two stories below the top.
b Three stories below the top
c Four stories below the top.
d none of the above
Physics
1 answer:
eduard2 years ago
4 0

Answer:

c Four stories below the top.

Explanation:

t = Time taken

u = Initial velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s_1=0\times t+\dfrac{1}{2}\times g\times 1^2\\\Rightarrow s_1=0.5g\ m

In two seconds

s_2=0\times t+\dfrac{1}{2}\times g\times 2^2\\\Rightarrow s_2=4\times 0.5g\\\Rightarrow s_2=4s_1

So, in two seconds the water balloon will fall 4 stories below the top

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Felix expends 100 W of power to clomb 10 meters in 20 seconds how much force does he exert
yulyashka [42]

Answer:

200 N

Explanation:

Power = work / time

Work = force × distance

Therefore:

Power = force × distance / time

100 W = F × 10 m / 20 s

F = 200 N

He exerted 200 N.

3 0
3 years ago
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A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constan
In-s [12.5K]

To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

\text{Initial Energy}=\text{Final Energy}

\frac{1}{2} mv^2=\frac{1}{2} kx^2

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

mv^2 = kx^2

v^2 = \frac{kx^2}{m}

v = \sqrt{\frac{kx^2}{m}}

Our values are,

m = 5.22*10^4kg

k = 4.58*10^5N/m

x = 32cm = 0.32m

Replacing our values we have,

v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}

v = 2.733*10^5m/s

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8 0
3 years ago
An object's starting position is at point A and its final position is at point B.
salantis [7]
B- we would need the time to fulfill the formula distance = speed x time
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What is the magnitude of the electric field at a point midway between a −5.0μC and a +5.8μC charge 8.4cm apart? Assume no other
Alex73 [517]

Answer:

Electric Field = E = 36.848 N/C

Explanation:

In accordance with Columb's law

E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084

= 36948.6961 x 10^-3 = 36.848 N/C

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E<br> 3.6 What force is needed to give a mass of<br> 20 kg an acceleration of 5 m/s??
luda_lava [24]

Explanation:

  • Mass(m)= 20kg
  • Acceleration (a)= 5m/s²
  • Force(F)= ?

We know that,

  • F=ma
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Hence, the needed force is 100N.

6 0
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