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GenaCL600 [577]
3 years ago
14

Which of your muscles push with the most force

Physics
1 answer:
noname [10]3 years ago
3 0

Answer:

Wouldn't it be the muscles in your arms or legs since they have to push the most while they have gravity acting upon them?

Explanation:

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What is the momentum of a 546,540 kg train that is travelling at 7.8 m/s​
lara [203]

p=mv so wouldn't u multiply them?

8 0
3 years ago
Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If
horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen 40 \mathrm{m}

Explanation:

We know, from faraday's law-

e m f=-N \frac{\Delta \Phi}{\Delta T}

and \Phi=B . A

So, as the height increases the velocity with which it will cross the ring will also increase. (v=\sqrt{2 g h})

Given

\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})

\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}

Now, from 40 \mathrm{cm}

V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}

From equation a and b we see that velocity when dropped from 40 \mathrm{m} is 10 times greater when height is 40 \mathrm{cm} so, emf will also be 10 times less as compared to when it has fallen 40 \mathrm{m}

4 0
3 years ago
For a wire has a circular cross section with a radius of 1.23mm.
Mila [183]

Answer:

5.731\times 10^{-5}\ m/s

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = 1.6\times 10^{-19}\ C

n = Conduction electron density in copper = 8.49\times 10^{28}\ electrons/m^3

v_d = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s

The drift speed of the electrons is 5.731\times 10^{-5}\ m/s

v_d=\dfrac{I}{neA}

From the equation we can see the following

v_d\propto \dfrac{1}{n}

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0
3 years ago
Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro
aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

Here,

Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0
3 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
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