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3 years ago

doc based on this lab: https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html

Physics

1 answer:

RoseWind [281]3 years ago

7 0

Answer:

hope it helps make brainlliest ty

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How fast will a ball be traveling if it falls for 5 seconds

Elenna [48]

Gravity accelerates you at 9.8 meters per second per second. After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.

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4 years ago

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

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3 years ago

An object is dropped from a tower, 400 ft above the ground. The object's height above ground t seconds after the fall is s(t)

Paul [167]

Answer: v= 160ft/s

a=32ft/s^2 constant

Explanation:

s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.

0=400-16t^2

400=16t^2

25=t^2

t=5s

ds/dt=v(t)=0-32t

dv/dt=a(t)=-32 constant(gravity)

v(t)=-32(5s)= -160ft/s negative sign is only showing direction

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3 years ago

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

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3 years ago

evaluate the force of a skateboard exerts on you if your mass is 60kg and you push on the skateboard with a force of 60N

Sav [38]

Every action has an equal or opposite reaction.
You weigh 60kg
<span>So your acceleration is 6N / 60 kg = 0.1m/s^2</span>

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3 years ago