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3 years ago

Feeling slightly uncomfortable during exercise

1 answer:

6 0

Feeling slightly uncomfortable during exercise is completely normal and should be expected. If there is unbearable pain however, the exerciser should stop and consult a doctor.

I hope this helped!

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A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const

Evgesh-ka [11]

Answer:

Both objects travel the same distance.

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

$v^2=u^2+2as$

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

$0= u^2-2as_{1}$

$s_{1}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{1}=\dfrac{8}{F}$

Now, For second object

Using equation of motion

$v^2=u^2+2as$

Put the value in the equation

$0= u^2-2as_{2}$

$s_{2}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$F = ma$

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{2}=\dfrac{8}{F}$

Hence, Both objects travel the same distance.

6 0

3 years ago

Using a simple machine, a student is able to lift a 500N weight by applying only 100N. What is the mechanical advantage of the s

Vanyuwa [196]

Answer:

The answer is 5

Explanation:

4 0

4 years ago

What is the force of gravity on a 24.0 kg table sitting on the earth

marysya [2.9K]

Answer:

Explanation:

4 0

3 years ago

A force of 6 N acts on a 33 kg object for 9 seconds. What is the objects change in velocity?

GuDViN [60]

Acceleration = change in velocity/time
By F = ma,
6 = 33 x change in velocity / 9
change in velocity = +1.636 m/s

4 0

3 years ago

A ships initial position is 25 km east of the origin. What is the ship’s final position if it sails at a constant velocity of 42

Evgen [1.6K]

Answer:

88km E

Explanation:

25 km is added already

then if we take the formula of speed(V)

V=d/t

and rearrange it to subject d (distance)

d= V x t

we get

d= 42km/h x 1.5hrs

we get

63 km

now if we add the already travelled 25 km to this additional distance

63km + 25km = 88km

Thus the final distance traversed by the ship from its origin is

88km E

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Hope this helps

7 0

3 years ago