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3 years ago

8

What injuries could occur due to incorrect movement and handling of equipment with any sport or activity? (Select ALL that apply

) Strains/sprains Dehydration Tears to muscles, ligaments, tendons Hypothermia

1 answer:

NeTakaya3 years ago

4 0

Sprains/Strains, tears to muscles, tendons ligaments, not hypothermia bc thats when you get really cold and not dehydrated

You might be interested in

Can you have zero displacement and nonzero average velocity? Zero displacement and nonzero velocity? Illustrate your answers on

Svetlanka [38]

a) Not possible

b) Yes, it's possible (see graph in attachment)

Explanation:

a)

The average velocity of a body is defined as the ratio between the displacement and the time elapsed:

$v=\frac{\Delta x}{\Delta t}$

where

$\Delta x$ is the displacement

$\Delta t$ is the time elapsed

In this problem, we want to have zero displacement and non-zero average velocity. From the equation above, we see that this is not possible. In fact, if the total displacement is zero,

$\Delta x = 0$

And therefore as a consequence,

$v=0$

which means that the average velocity is zero.

B)

Here we want to have zero displacement and non-zero velocity. In this case, it is possible: in fact, we are not talking about average velocity, but we are talking about (instantaneous) velocity.

On a position-time graph, the instantaneous velocity is the slope of the graph. Look at the graph in attachment. We see that the position of the object first increases towards positive value, then it decreases (the object starts moving backward), then becomes negative, then it increases again until returning to the original position, x = 0.

In all of this, we notice that the total displacement of the object is zero:

$\Delta x = 0$

However, we notice that the instantaneous velocity of the object at the various instants is not zero, because the slope of the graph is not zero.

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

3 0

3 years ago

Which statement correctly compares ultraviolet light to visible light? Ultraviolet light has both a lower frequency and longer w

Effectus [21]

The correct statement is

Ultraviolet light has both a higher frequency and a higher radiant energy than visible light.

because ultraviolet light has wavelength smaller than the visible light hence has a greater frequency as compared to visible light. (frequency is inversely related to wavelength. hence smaller the wavelength , greater will be the frequency)

we also know that the radiant energy is directly proportional to the frequency. hence greater the frequency , greater will be the radiant energy.

Since the frequency is greater for ultraviolet light , it radiant energy is also greater

7 0

3 years ago

Read 2 more answers

Lab: Thermal Energy Transfer What is the purpose of the lab, the importance of the topic, and the question you are trying to ans

VladimirAG [237]

Answer:

it's important because it shows how thermal energy transforms or continues to be all around us in everything

6 0

3 years ago

A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 324 meters. Assuming no air resist

Elanso [62]

Answer:

8.13secs

Explanation:

From the question weal are given

Height H =324m

Required

time it takes to drop t

Using the equation of motion

H = ut + 1/2gt²

Substitute the given values

324 = 0(t)+1/2(9.8)t²

324 = 1/2(9.8)t²

324 = 4.9t²

t² =324/4.9

t² = 66.12

t = √66.12

t = 8.13secs

Hence the time taken to drop is 8.13secs

4 0

3 years ago

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

6 0

3 years ago