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alexdok [17]
3 years ago
8

Which of the following is not a way to make exercising on a cold day safer?

Physics
2 answers:
IRISSAK [1]3 years ago
4 0

Answer:

D.

Explanation:

I think its D. because Tea during a cold workout isnt safe at all. Drinking tea while working out adds weight and you just simply dont need tea when working out.. Hope this helps man :)

dsp733 years ago
4 0

C. You shouldn't exercise while drinking hot tea because you can spill it and burn yourself especially while on a treadmill.

If this helped please mark me as brainliest!

\left[\begin{array}{ccc}Thanks,\\JustSomeIdiot\end{array}\right]

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MatroZZZ [7]
Plyometric does indeed help strengthen your bones
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2 years ago
A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe
Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

I_R=\frac{1}{12}ML^2

where

M=3.30\cdot 10^{-2} kg is its mass

L = 0.450 m is its length

Substituting,

I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2

The moment of inertia of the two rings at the beginning is

I_r = 2mr^2

where

m = 0.200 kg is the mass of each ring

r=5.20\cdot 10^{-2} m is their distance from the center of the rod

Substituting,

I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2

So the total moment of inertia at the beginning is

I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2

The initial angular velocity of the system is

\omega_1 = 35.0 rev/min

The angular momentum must be conserved, so we can write:

L=I_1 \omega_1 = I_2 \omega_2 (1)

where I_2 is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

r=\frac{0.450 m}{2}=0.225 m

so the moment of inertia of the rings is

I_r=2(0.200)(0.225)^2=0.0203 kg m^2

and the total moment of inertia is

I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2

Substituting into (1), we find the final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

I_2 = I_R = 5.57\cdot 10^{-4}kg m^2

So using again equation of conservation of the angular momentum:

L=I_1 \omega_1 = I_2 \omega_2

We find the new final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min

7 0
2 years ago
What is the atomic number of a sodium atom that has 11 protons and 12 neutrons
Ede4ka [16]
The Atomic number for sodium is 11 cause it contains 11 protons. I hope this helps..
8 0
2 years ago
A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli
Gre4nikov [31]

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble m_1=41gram=0.041kg

Velocity of marble v_1=2.30m/sec

Its collides with other marble of mass 25 gram

So mass of other marble m_2=25gram=0.025kg

Second marble is at so v_2=0m/sec

We have to find the velocity of second marble

From momentum conservation we know that

m_1v_1+m_2v_2=(m_!+m_2)v, here v is common velocity of both marble after collision

So 0.041\times 2.30+0.025\times 0=(0.041+0.025)v

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

6 0
2 years ago
A 6.00 A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5×1028 free elect
navik [9.2K]

Answer:

Explanation:

Current, I = 6 A

diameter of wire, d = 2.05 mm

number of electrons per unit volume, n = 8.5 x 10^28

If the diameter is doubled,

The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is  one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.  

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