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2 years ago

What is the benefit of cooling down after exercise?

Physics

1 answer:

d1i1m1o1n [39]2 years ago

3 0

Answer:

3 benefits of a restrict afterwards a practice Brings essence rate back off and slows respiring Muscles return back to optimum motionless distance Prevents ancestry from combining in lower limits Cool down stretches are used to weaken courage rate and respiring rates afterwards exercise. It can help cause your soul rate back off to situated rate.

Explanation:

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(a) Calculate the height of a cliff if it takes 2. 35 s for a rock to hit the ground when it is thrown straight up from the clif

Natali [406]

(a) The height of the cliff will be 8.26 meters.

(b) The time would it take to reach the ground will be 0.717 sec.

<h3>What is velocity?</h3><h3 />

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component.

(a) The height of the cliff will be 8.26 meters.

According to Newton's second equation of motion

$\rm H =ut-\frac{1}{2} gt^2 \\\\ \rm H =8\times 2.35-\frac{1}{2} 9.81 (2.35)^2\\\\\rm H =8.16 \; m$

Hence The height of the cliff will be 8.26 meters.

(b)The time would it take to reach the ground will be 0.717 sec.

We must have the final velocity to find the time so;

$\rm v^2=u^2+2gh\\\\ \rm v^2=8^2+2\times 9.81 \times 8.6 \\\\ \rm v= \sqrt{8^2+2\times 9.81 \times 8.6}\\\\\rm v=15.03 \;m/sec$

According to Newton's third equation of motion ;

$\rm v=u-gt \\\\ \rm t=\frac{v-u}{g} \\\\ \rm t=\frac{15.03-8}{9.81} \\\\ \rm t=0.717 sec.$

Hence the time would it take to reach the ground will be 0.717 sec.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

3 0

3 years ago

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

$261.3=0.3\times 9.8\times h$

$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

4 0

3 years ago

A rock is rolling down a hill, and it’s halfway down. Would it have both Kintectic engery and Potenial engery? PLEASE HELP

dimaraw [331]

Answer:

Only kinetic.

Explanation:

Potential energy means it has the potential to move. Not something already in motion.

6 0

4 years ago

A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.

ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

$P = VI$

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

$I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\$

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

$V = IR$

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

$R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega$

Therefore, the resistance of the bulb is 484 Ω

3 0

3 years ago

Read 2 more answers

How will the motion of the arrow change after it leaves the bow?

Pavel [41]

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

4 0

3 years ago