What kind of exercise should you do when you're cooling down after an
intense workout?

Answers

Answer 1
Answer:

Answer:

Planks?

Explanation:

It's kinda resting

Answer 2
Answer:

Answer:

weights

walking

stretching

etc.


Related Questions

A local meteorologist reports the day’s weather. "Currently sunny outside, 34°F. Skies will become overcast later this afternoon, as temperatures drop to 25°F, with windy conditions out of the north at 10–15 miles per hour. Radar indicates 2–3 inches of snow expected to fall later tonight.” Which information is qualitative? These are non-numerical, descriptive data. These are numerical data that have been measured. “sunny” “25°F” “2–3 inches of snow” “10–15 miles per hour”
A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pull the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg. -Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well. -Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.
The temperature within a thin plate with thermal conductivity of 10 W/m/K depends on position as given by the following expression: TT=(100 K)????????−xx2/????????xx 2cos�yy/????????yy�+300 K Where, Lx = 1 m, and Ly = 2 m. At the point (0.4 m, 1 m), find: a. The magnitude of the heat flux b. The direction of the heat flux
Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many more turns should be on the secondary compared with the primary?
A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Two 3.7 microCoulomb charges are 0.8 m apart. How much energy (in milliJoule) went into assembling these two charges? Enter a number with one digit behind the decimal point.

Answers

Answer:

Explanation:

The potential energy of a system of two charges is given by the expression

(K* Q_1* Q_2)/(R)

Q₁ and Q₂ are two charges and R is distance between the charges.

Given Q₁ = Q₂ = 3.7 X 10⁻⁶ , R = .8 and K = 9 x 10⁹

Putting these values in the equation we have,

[tex](9* 10^9* 3.7*10^(-6) 3.7* 10^(-6))/(.8)[/tex]

Potential energy = 154.01 x 10⁻³ J

This energy have been spent to bring these repelling two charges at this close distance . The energy spent have been stored as potential energy here which has been calculated.

1. A bicyclist starts at 2.5 m/s and accelerates along a straight path to a speed of 12.5 m/s ina time of 4.5 seconds. What is the bicyclist's acceleration to the nearest tenth of a m/s??

Answers

Answer:

2.2m/s

Explanation:

a=v-u/t

12.5-2.5/4.5=2.222

~2.2m/s

Someone claps his or her hands is an example of… motion energy to sound energy sound energy to potential energy motion energy to radiant energy

Answers

motion energy to sound energy i think

An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge −Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. Show that the electric field inside this atom is : Ein=Ze4πϵ0(1r^2−rR^3). b. What is the electric field at the surface of the atom? Is this the expected value? Explain.c. A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field at r = R/2?

Answers

Answer:

Part a)

E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))

Part b)

E = 0

Yes it is the expected value of electric field at the surface of an atom

Part c)

E = 4.64 * 10^(13) N/C

Explanation:

Since negative charge of electrons in uniformly distributed in the atom while positive charge is concentrated at the nucleus

So the electric field due to positive charge of the nucleus is given as

E = (kq)/(r^2)

E_1 = (Ze)/(4\pi \epsilon_0 r^2)

now charge due to electrons inside a radius "r" is given as

q = (-Ze r^3)/(R^3)

now we will have electric field given as

E_2 = ((-Zer^3)/(R^3))}{4\pi\epsilon_0 r^2}

now net electric field is given as

E = E_1 + E_2

E = (Ze)/(4\pi \epsilon_0 r^2) - (Zer)/(4\pi \epsilon_0 R^3)

E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))

Part b)

At the surface of an atom

r = R

E = 0

Yes it is the expected value of electric field at the surface of an atom

Part c)

If Z = 92

R = 0.10 nm

r = (R)/(2)

so we will have

E = 92(1.6 * 10^(-19)) * (9 * 10^9)((4)/(R^2) - (1)/(2R^2))

E = (4.64 * 10^(-7))/((0.10 * 10^(-9))^2)

E = 4.64 * 10^(13) N/C

11.
A current of 67 amps runs through a resistor of 37 ohms, how much voltage is lost?

Answers

You divide them. 67/37 is 1.5 and so you subtract it with the 67 and multiple it by the coherent integer from the multiplication and you would get 20 volts lost roughly.

Answer: 20 volts

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

Answers

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) h*(1 - 1/2 g * h/v_0^2)

b)h = v_0^2/ g

c)h = v_0^2/ g

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t

where:

  • y = height at time t
  • y0 = initial height
  • v0 = initial velocity
  • a = acceleration
  • t = time
  • v = velocity

a) When the balls collide, h1 = h2. Then,

h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0

Replacing in the equation of the height of the first ball:

h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)

b)  that the balls collide at t = h/v0. Then:

h/ v_0 = -v_0/-g\nh = v_0^2/ g

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

h = v_0^2/ g

See more about velocity at brainly.com/question/862972