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3 years ago

HELP is it impossible to experience free fall near the earths surface? why?

1 answer:

3 0

It’s not impossible because the gravitational pull from the earths gravity from the atmosphere will pull you in just like all other planets that’s why you can fall at really fast speeds from the air.

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A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr

Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = $\frac{9\times 10^9\times3\times10^{-9}}{d^2}$

Field due to charge = 4 X 10⁻⁹ C

$E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$

These two fields will be equal and opposite to make net field zero

$\frac{9\times 10^9\times3\times10^{-9}}{d^2}$ = $[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$ [/tex]

$\frac{3}{d^2} =\frac{4}{(2-d)^2}$

$\frac{2-d}{d} =\frac{2}{1.732}$

d = 0.928

5 0

3 years ago

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in

iVinArrow [24]

Answer:

The value is $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 14\ psla$

The initial temperature is $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

The final temperature is $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

$PT^{\frac{\gamma}{1- \gamma} } = Constant$

=> $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas $\gamma = \frac{5}{3}$

$14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=> $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=> $P_2 = 40.54 \ psla$

8 0

2 years ago

What shape is the Milky way galaxy

gizmo_the_mogwai [7]

It appears to be a <span>spiral shape. </span>

5 0

3 years ago

Read 2 more answers

You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w

Vesna [10]

Answer:

$\ \text{m/s}$

Explanation:

$u_1$ = Velocity of one lump = $3x+3y-3z$

$u_2$ = Velocity of the other lump = $-4x+0y-4z$

m = Mass of each lump = $30\ \text{g}$

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

$mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}$

The velocity of the stuck-together lump just after the collision is $\ \text{m/s}$ .

4 0

2 years ago

Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.

ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0

2 years ago