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Sunny_sXe [5.5K]
3 years ago
14

How much does a 44 kg child weigh on Earth?​

Physics
2 answers:
KIM [24]3 years ago
5 0

Answer:

96.8 or 97 pounds if you need to round

Explanation:

To convert kilograms to pounds you multiply the kilogram with 2.2. So 44kg × 2.2 equals to 96.8 but if you round it turns to 97.

Akimi4 [234]3 years ago
4 0
Well if the child weighs 44 kg then 44 kg. Do you mean if they weigh 44 kg in space to earth?
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Find the resultant of two components, 3km west and 4 km south. How do I solve this using the Pythagorean theorem?
Rasek [7]

Answer:5

Explanation:

a=3

b=4

c=√(a²+b²)

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3 0
3 years ago
If a ball is dropped and takes 6.5 seconds to hit the ground below. How far did it fall? Remember in free fall g= -9.8 m/s/s. Wh
vovikov84 [41]

Answer:

See the answers below.

Explanation:

To solve this problem we must use the following equation of kinematics.

y=y_{o}+v_{o}*t+\frac{1}{2}*g*t^{2}  \\

where:

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t = time = 6.5 [s]

g = gravity acceleration = 9.8 [m/s²]

Now replacing:

y-y_{o}=0 +\frac{1}{2} *9.8*6.5^{2}\\y-y_{o}=207 [m]

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3 0
3 years ago
A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-
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Answer:

5.25\cdot 10^{40} kg m^2/s

Explanation:

The angular momentum of the pulsar is given by:

L=m\omega r^2

where

m=2.8\cdot 10^{30} kg is the mass of the pulsar

r = 10.0 km = 1\cdot 10^4 m is the radius

\omega is the angular speed

Given the period of the pulsar, T=33.5\cdot 10^{-3} s, the angular speed is given by

\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s

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L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s

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