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melisa1 [442]
3 years ago
5

If the total time is 120 seconds and the total distance is 320 meters. Calculate

Physics
1 answer:
RUDIKE [14]3 years ago
3 0

Answer:

S=D/T

320/120= 2.66 miles

Explanation:

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A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

  • <em>momentum of the ball, P = 0.9 kgm/s</em>
  • <em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

Ft = \Delta P

where;

Ft is impulse

\Delta P is change in momentum

The time of motion of the ball is calculated as follows;

t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

7 0
2 years ago
According to newtons third law, the two forces in a force pair act
aliina [53]
B- Same force


It’s b because force always acts in equal but opposite pairs.
6 0
3 years ago
From the circuit above, predict which bulb (or bulbs) will be the brightest. Why do you think that?
inysia [295]

Answer:

the middle

Explanation:

the left one bulb gets power from the outher bulb

the one on right has more bulbs

7 0
3 years ago
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A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati
JulsSmile [24]
m=852 \ kg \\ h=3,5 \ m \\ g=9,8 \ m/s^2 \\ \boxed{P_e-?} \\ \bold{Solving:} \\ \boxed{P_e=m \cdot g \cdot h} \\ P_e=852 \ kg \cdot 9,8 \ m/s^2 \cdot 3,5 \ m =8 \ 349,6 \ N \cdot 3,5 \ m \\ \Rightarrow \boxed{P_e=29 \ 223,6 \ J}
7 0
3 years ago
The time period T of a simple pendulum is given by the relation
Vanyuwa [196]

Answer:

T^2 \propto L

Explanation:

The period of a simple pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

From this equation we can write

T\propto \sqrt{L}\\T\propto \frac{1}{\sqrt{g}}

Taking the square of this equation, we get:

T^2 = (2\pi)^2 \frac{L}{g}

So we see that T^2 is proportional to L and inversely proportional to g. So, we can write:

T^2 \propto L\\T^2 \propto \frac{1}{g}

So the only correct option is

T^2 \propto L

5 0
3 years ago
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