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4 years ago

Swinging a golf club or baseball bat are examples of ______________ stretching.

Physics

1 answer:

kari74 [83]4 years ago

6 0

Answer:

Static stretching is the answer.

Explanation:

Static stretching is the most common form that greatly improves flexibility. However, static stretches does little to contract the muscles needed to generate powerful golf swings. Dynamic stretches help improve your range of motion while reducing muscle stiffness.

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If atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm at 298 k, how much oxygen will be released from 3.30 l of wa

xxMikexx [17]

At a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:

<span>S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M
</span>
At 0.890 atm,
<span>S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M</span>
<span>
If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows</span>
<span>
3.30 L x (0.000273 mol / L) = 0.0012012 mol</span>

3.30 L x (0.00024297 mol / L) = 0.001069068 mol

0.0012012 mol - 0.001069068 mol = 0.000132 mol

6 0

4 years ago

Help me please savior

kolbaska11 [484]

Answer:

-16°C

Explanation:

PV = nRT

V and n are constant.

P / T = P / T

(2 atm + 1 atm) / (266 K) = (1.9 atm + 1 atm) / T

T = 257.1 K

T = -16°C

7 0

3 years ago

Two cats are sitting on a 30 m hill. One cat

V125BC [204]

Answer:

Cat with 6 kg sitting on a 30 meter hill has greater potential energy.

Explanation:

gravitational potential energy (J) = mass (kg) × gravitational field × height (m)

gravitational field in Earth is 10 N/kg = <u>fixed unit</u>

<h3 /><h3><u>Given</u>:</h3>

1st Cat: 4 kg

2nd Cat: 6 kg

Both of their height is 30 meters.

<h3><u>Solve for g.p.e</u>:</h3>

1st cat = mgh = 4 × 10 × 30 = 1200 Joules

2nd cat = mgh = 6 × 10 × 30 = 1800 Joules

3 0

2 years ago

A small car with a mass of 800kg moving with a velocity of 27.8 m/s. The car stops at a yellow light in 3.9 seconds. What force

raketka [301]

Answer:

F = 5702.56 N

Explanation:

Given that,

Mass of a small car, m = 800 kg

Initial speed of the car, u = 27.8 m/s

Final speed, v = 0

Time, t = 3.9 s

We need to find the force did it take for the car to stop.

The force acting on an object is given by :

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{800\times (0-27.8)}{3.9}\\\\F=-5702.56\ N$

So, the magnitude of force acting on the car to stop is 5702.56 N.

4 0

3 years ago

An AC voltage source has an output of ∆V = 160.0 sin(495t) Volts. Calculate the RMS voltage. Tries 0/20 What is the frequency of

kupik [55]

Answer:

RMS voltage is 113.1370 V

frequency is 780.685 Hz

voltage is −158.66942 V

maximum current is 2.9739 A

Explanation:

Given data

∆V = 160.0 sin(495t) Volts

so Vmax = 160

and angular frequency = 495

time t = 1/106 s

resistor R = 53.8 Ω

to find out

RMS voltage and frequency of the source and voltage and maximum current

solution

we know voltage equation = Vmax sin ωt

here Vmax is 160 as given equation in question

so RMS will be Vmax / √2

RMS voltage = 160/ √2

RMS voltage is 113.1370 V

and frequency = angular frequency / 2π

so frequency = 497 / 2π

frequency is 780.685 Hz

voltage at time (1/106) s

V(t) = 160.0 sin(495/ 108)

voltage = −158.66942 V

so current from ohm law at resistor R 53.8 Ω

maximum current = voltage max / resistor

maximum current = 160 / 53.8

maximum current = 2.9739 A

8 0

3 years ago