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fomenos
3 years ago
10

How many exercises are there? ​

Physics
2 answers:
trasher [3.6K]3 years ago
7 0

Answer:

There are 4 exresise ;endurance, strength, balance, and flexibility. Each one has different benefits.

Explanation:

Endurance activities, often referred to as aerobic, increase your breathing and heart rates.

Your muscular strength can make a big difference. Strong muscles help you stay independent and make everyday activities feel easier, like getting up from a chair, climbing stairs, and carrying groceries.

Balance exercises help prevent falls, a common problem in older adults that can have serious consequences.

Stretching can improve your flexibility. Moving more freely will make it easier for you to reach down to tie your shoes or look over your shoulder when you back your car out of the driveway.

hope this can help you...

aivan3 [116]3 years ago
4 0
There are 4 exercises! I hope this helps :)
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Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
Explain what happens to light when it is refracted at the surface of water.
stepan [7]
Light will make the object appear “broken” or in an irregular shape.

Refraction is the change in direction of waves.
6 0
3 years ago
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A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
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Klio2033 [76]

Answer:

So I never really knew you

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Felt we could really do this

But really I was foolish

Hindsight, it's obvious

Explanation:

3 0
3 years ago
Please help the question is the picture below.
balu736 [363]

Its 30 kg cause I got 30 kg dude.

8 0
3 years ago
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