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3 years ago

PLEASE ANSWER ASAP!!! ITS DUE IN 10 MINUTES!!!!!!

Physics

1 answer:

forsale [732]3 years ago

5 0

Explanation:

2. Aerobic exercise. ...

Strength building. ...

Balance Training. ...

Endurance. ...

Flexibility. ...

Moderate intensity exercise. ...

Vigorous exercise.

3.Brisk walking.

Running.

Jogging

Bear crawls.

Swimming.

Water aerobics.

Cycling/bicycling.

4. Running or Jogging. ...

Outdoor Cycling. ...

Walking.

5. The type of exercise to build muscle is strength training, although cardiovascular activity can also provide benefits.

Hope it helps..

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Which is heavier, 1 m3 of steel or 1 m3 of aluminium?

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Answer:

Steel is almost 2.9 times heavier the aluminium.

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135,000 kilometers = how many miles?

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3 years ago

How many protons does Silicone have A.2 B.14 C.28 D.28.08

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3 years ago

A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ

lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

$U=\frac{Kq_{1}q_{2}}{r}$

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

$U = U_{1,2}+U_{2,3}+U_{3,1}$

$U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}$

$U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}$

U = - 2.7 x 10^-6 J

3 0

3 years ago

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

lesya [120]

Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

Explanation:

Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

$q_2=7.13\times 10^{-6}\ C$

$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

5 0

3 years ago