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Crank
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
Physics
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
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Why is a football firm when it is inflated to its proper pressure
Molodets [167]

Answer:

Proper Inflation and Feel When inflating your ball, you can use either a hand pump or an air pump equipped with a gauge that gives readings in pounds per square inch, also called psi. Footballs used in the NFL are inflated to 13 psi, but a proper range can fall between 12.5 and 13.5 psi, according to Wilson Sporting Goods.

Explanation:

8 0
3 years ago
Difference between relaxation time and collision time?
Black_prince [1.1K]
For the answer to the question above, let us first start with relaxation time. it is the absence of an external electric field, the free electrons in a metallic substance will move in random directions so that the resultant velocity of free electrons in any direction is equal to zero. While the Collision time it is<span> the mean </span>time<span> required for the direction of motion of an individual type particle to deviate through approximately as a consequence of </span>collisions<span> with particles of type.</span>
6 0
3 years ago
Read 2 more answers
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
3 years ago
8
Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0
3 years ago
12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and
Nimfa-mama [501]

Answer:

I think 9.5

Explanation:

............

6 0
3 years ago
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