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3 years ago

Why are overuse injuries particularly frustrating set-backs?

Physics

2 answers:

olganol [36]3 years ago

7 0

Because they are caused by your exercise

anastassius [24]3 years ago

4 0

Answer:

Because they are caused by the excess exercise and lack of rest to recover the body.

Explanation:

Over use injuries can be say that the inflammation of the connecting tissues which attaches the bones to the muscles. They can be occur if someone does not modify his or her workout after a break.

These injuries particularly are frustrating set backs because you are not give a much time of rest to your body to recover from these injuries.

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The change in momentum of an object is equal to the Question 4 options: Force acting on it times its velocity. impulse acting on

lesya [120]

Answer:

impulse acting on it

Explanation:

The impulse is defined as the product between the force applied to an object (F) and the time interval during which the force is applied ( $\Delta t$ ):

$I=F\Delta t$

We can prove that this is equal to the change in momentum of the object. In fact, change in momentum is given by:

$\Delta p = m \Delta v$

where m is the mass and $\Delta v$ is the change in velocity. Multiplying and dividing by $\Delta t$ , we get

$\Delta p = m \frac{\Delta v}{\Delta t} \Delta t$

and since $\frac{\Delta v}{\Delta t}$ is equal to the acceleration, a, we have

$\Delta p = ma \Delta t$

And since the product (ma) is equal to the force, we have

$\Delta p = F \Delta t$

which corresponds to the impulse.

5 0

3 years ago

Help me please, i need it fast

s2008m [1.1K]

Answer:

2) C. (x - 3)² + (y + 2)² = 25

5) x² + y² - 8x - 16y + 54 = 0

6) x² + y² - 10x - 12y + 36 = 0

Explanation:

center of circle = 3, -2

x1, y1

end point of circle = 7, 1

x2, y2

the equation of a circle is Pythagorean theorem

x² + y² = r² (where r is the radius of a circle)

distance between points

(x2 - x1)² + (y2 - y1)² = r²

(7 - 3)² + (1 - (-2))² = r²

r² = 25

therefore the equation to the circle is

(x - 3)² + (y + 2)² = 25

=========================================

write the general form of a circle with the center (4,8)

and containing the point (-1, 7)

distance between points

(x2 - x1)² + (y2 - y1)² = r²

(-1 - 4)² + (7 - 8)² = r²

r² = 26

(x - 4)² + (y - 8)² = 26

(x - 4)(x - 4) + (y - 8)(y - 8) = 26

x² - 8x + 16 + y² - 16y + 64 -26 = 0

x² + y² - 8x - 16y + 54 = 0

=========================================

find the general form of a circle with center (5,6)

and tangent to the y-axis.

center (5,6)

h, k

radius = r²

r = 5

(x - h)² + (y - k)² = r²

(x - r)² + (y - k)² = r²

(x - 5)(x - 5) + (y - 6)(y - 6) = r²

x² - 10x + 25 + y² - 12y + 36 = 25

x² + y² - 10x - 12y + 36 = 0

=========================================

8 0

3 years ago

What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan

kramer

I think the question should be the below:

What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answer is the below:

Acceleration .. a = (-) ω² x 
(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) 

x (max) = a(max) /ω² 

x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)

5 0

3 years ago

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.

Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA\dfrac{dB}{dt}$

For N = 1

$\epsilon=A\dfrac{dB}{dt}$

We need to calculate the current

Using formula of current

$i=\dfrac{\epsilon}{R}$

Put the value of emf

$i=\dfrac{A\dfrac{dB}{dt}}{R}$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA\dfrac{dB}{dt}$

Then the current would be

$i'=\dfrac{\epsilon'}{NR}$

$i'=\dfrac{NA\dfrac{dB}{dt}}{NR}$

$i'=\dfrac{A\dfrac{dB}{dt}}{R}$

$i'=i$

Hence, The current would be same in both situation.

4 0

3 years ago

A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?

lesantik [10]

<h3>Answer;</h3>

1600 years

<h3>Explanation;</h3>

Half life is the time taken for a radioactive isotope to decay by half of its original amount.
We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

1/4 = 1 × (1/2)^n

1/4 = (1/2)^n

n = 2

Thus; 3200 years is equivalent to 2 half lives.

Hence, the half life of radium-226 is 1600 years

3 0

4 years ago