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3 years ago

What forces are acting when you try to crack an egg in your palm(will give brainliest)

Physics

1 answer:

Aleks [24]3 years ago

4 0

Answer:

Kinetic energy in your palm

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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5

schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0

3 years ago

Newton's first law of motion is the law of inertia. What does it state?

vladimir2022 [97]

Answer:

b is the right answer

Explanation:

<h3>hope helps!!</h3>

5 0

3 years ago

The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands

Ghella [55]

Answer:

Pressure on both feet will be $500N/m^2$

Explanation:

Weight of the person F = 500 N

Area of foot print $A=0.5m^2$

Area of both the foot $A=2\times 0.5=1m^2$

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure $P=\frac{F}{A}$

$P=\frac{500}{1}=500N/m^2$

So the pressure on both feet will be $500N/m^2$ when person stands on both feet.

7 0

3 years ago

A Michelson interferometer uses red light with a wavelength of 656.45 nm from a hydrogen discharge lamp. Part A How many bright-

kati45 [8]

Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

$\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}$

Here,

$\Delta L$ = is the distance moved by the mirror M

$\lambda$ is the wavelenght of the light used.

$\Delta L$ = 0.017m

$\lambda = 656.45*10^{-9}m$

$\Delta m = \frac{0.017}{\frac{656.45*10^{-9}}{2}}$

$\Delta m = 51793.72$

Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7

3 0

3 years ago

Tsunami waves are caused by earthquakes in the ocean floor. As the waves approach the shore, they usually grow in height. When t

aksik [14]

Answer:

LETTER A. AMPLITUDE

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3 years ago

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