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3 years ago

Does increasing impulse means that theres less momentum?

1 answer:

4 0

in every collision there will be always an impulse I which represents change in momentum and is always constant. Thus, I = F x t = constant. Hence whenever, there is a change in impact ie force, there will also be equal change in time or vice versa. Thus, increae of time will decrease impact ie., force.

e.g., a person gets less injured on sand floor when compared to cemented floor when jumping from same height due to increase of time and decrease of force.

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An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

3 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago

A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

Yuri [45]

W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.

5 0

3 years ago

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Which will result in a greater increase in kinetic energy, doubling the velocity or doubling the mass.

Naddik [55]

Kinetic energy is the energy possessed by a body while in motion. It is calculated by 1/2mv², where m is the mass of the body and v is the velocity.
Therefore, kinetic energy is dependent on both mass of the body and the velocity. An increase in mass increases the kinetic energy, an increase in velocity also increases kinetic energy of the body. Thus, doubling the mass and doubling the velocity will both increase the kinetic energy of the body.

6 0

3 years ago

Write down any four points that should be considered during household wiring

Shalnov [3]

1. Safety equipment is available
2. Person attempting the task has some general knowledge about wiring
3. Not All Cable is Color-Coded
Cable-sheath color coding started in 2001 and is still voluntary. If you have older wiring, don’t assume it complies with the current color coding. However, most manufacturers now follow the standard color code.
4. Stranded wire is more flexible than solid. If you’re pulling wire through conduit, stranded wire makes it easier to get around corners and bends in the conduit. However, if the situation requires pushing wires through conduit, you’ll want to use solid wire.

8 0

3 years ago

Read 2 more answers