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3 years ago

It is a general exercise that we do before practicing sports activities worm-up or cool down

Physics

2 answers:

ludmilkaskok [199]3 years ago

8 0

Answer:

We do warm-ups before we exercise. We do cool-downs after we exercise.

Explanation:

Warming up can help you get ready to do the actual exercise, so you are prepared.

Montano1993 [528]3 years ago

3 0

It is called a warm-up.

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A skydiver is preparing to jump out of a plane that is flying at a height of 2200 m. If the skydiver has a mass of 78 kg, what i

Illusion [34]

Answer:

1,681,680 J.

Explanation:

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3 years ago

If a certain mass has its velocity changed from 6.00 m/s to 7.50 m/s when a 3.00 n force acts for 4.00 seconds, find the mass of

shutvik [7]

By definition we have that the force for time is equal to the product of the mass for the change in speed.
We have then that
F * (delta t) = m * (delta v)
Clearing the mass
m = (F * (delta t)) / (delta v)
Substituting the values
m = ((3.00) * (4.00)) / (7.50-6.00) = 8
answer
The mass of the moving object is 8Kg

3 0

4 years ago

Read 2 more answers

What is a purpose in flexibility <br>

icang [17]

Answer:

To prevent from getting injured.

8 0

3 years ago

Read 2 more answers

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

Yuki888 [10]

Answer: $F=3\times 10^{-5} N$

Explanation:

Given

$I_1=2 Amps$

$I_2=0.75 Amps$

Length of each wires $\left ( L\right )=5 m$

Distance between wires $\left ( r\right )=5 cm$

Force per unit length = $\frac{\mu_0I_1I_2}{2\pi r}$

$F'=\frac{2\times 10^{-7}\times 2\times 0.75}{2\pi \times 0.05}$

$F'=6\times 10^{-6}$

Force for $L=5 m$

$F=3\times 10^{-5} N$

6 0

3 years ago

A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei

Mariana [72]

Answer:

the speed of the car at the top of the vertical loop $v_{top} = 2.0 \sqrt{gR \ \ }$

the magnitude of the normal force acting on the car at the top of the vertical loop $F_{N} = 1.47 \ \ N$

Explanation:

Using the law of conservation of energy ;

$mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R$

$v_{top} = 2.0 \sqrt{gR \ \ }$

The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

$F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\$

$F_{N} = 1.47 \ \ N$

4 0

4 years ago