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4 years ago

John stretched while his muscles were cold. Which negative effect could occur?

Physics

2 answers:

Ede4ka [16]4 years ago

5 0

Answer:

The answer for Plato is C.

Genrish500 [490]4 years ago

3 0

<h3>Answer;</h3>

C) He could injure or pull a muscle.

<h3>Explanation;</h3>

When muscles are stretched the muscle fibers are temporary lengthened. Muscles have a unique ability to undergo contraction and lengthening since they are elastic.
When the muscles are warm they are more elastic, therefore, warming up the muscles is very important for the purpose of avoiding injuries. Ir is important to engage in moderate cardiovascular warm-up prior to stretching, which increases the blood flow to the active and thus avoiding injuries to the muscles.
Stretching muscles when they are cold may bring injuries or cause muscle pulls, which are painful.

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A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR.

Snowcat [4.5K]

A) The vertical component of velocity v is taking the rock to a height

Vertical component = $vsin\theta$
The time taken to reach maximum height = $\frac{vsin\theta}{g}$
So total time of rocks flight = $\frac{2vsin\theta}{g}$
Range of rock is due to the horizontal component of velocity = $vcos\theta$
Range = $\frac{2*v*sin\theta*v*cos\theta}{g}$ = $\frac{2*v^2*sin\theta*cos\theta}{g}$
Maximum height = $\frac{g*t^2}{2}$ = $\frac{v^2*sin^2\theta}{2*g}$
Since range = maximum height
We have $\frac{2*v^2sin\theta*cos\theta}{g} = \frac{v^2*sin^2\theta}{2*g}$
$tan\theta = 4$
$\theta = 75.96^0$
So when angle of projection is $\theta = 75.96^0$ range is equal to maximum height reached.
b) We have range = $\frac{2*v^2*sin\theta*cos\theta}{g}$ = $\frac{2*v^2*sin2\theta}{g}$
Maximum of range is reached when $\theta = 45^0$
Maximum range = $\frac{2*v^2}{g}$
c) For range to be equal to maximum height only condition is $tan\theta = 4$ , it does not depend upon acceleration due to gravity and velocity. That angle is a constant.

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3 years ago

The 75.0 kg hero of a movie is pulled upward with a constant acceleration of 2.00 m/s2 by a rope. What is the tension on the rop

lutik1710 [3]

Use Newton's second law. The net force on the hero is

∑ F = T - m g = m (2.00 m/s²)

where

• T = tension in the rope

• m = 75.0 kg = mass of the hero

• g = 9.80 m/s² = acceleration due to gravity

Solve for T :

T = (75.0 kg) (g + 2.00 m/s²) = 885 N

7 0

3 years ago

ASAP pls answer right I will mark brainiest . All I know is 4. Is A

Sonja [21]

Answer:

Q1: B.2 Q2: B.Waxing crescent Q3: A.Waxing Gibbous

Explanation:

4 0

3 years ago

What was the most serious money problem faced by people in the early United States?

alexira [117]

♥B. There was no single medium of exchange.♥
Take a look at the saying "Not worth a Continental".
It referrers to the fact that currency issued by the Continental Congress was usually just a promise to pay.
Leaving your answer with ♥B♥

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3 years ago

A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the vel

spin [16.1K]

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

$m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s$

So Option d is correct one

5 0

3 years ago