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3 years ago

Which graph shows an object that is dropped?

Physics

1 answer:

Olegator [25]3 years ago

8 0

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the

Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

$\frac{F_E}{F_G}=\frac{qE}{mg}$ (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

$\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764$

Then, the gravitational force is 0.764 times the electric force on the particle

The acceleration of the particle is obtained by using the second Newton law:

$F_E-F_G=ma\\\\a=\frac{qE-mg}{m}$

you replace the values of all variables:

$a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}$

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0

3 years ago

You walk 45m west then 35m west

Nat2105 [25]

And walk another 25m west

4 0

3 years ago

What is the gauge pressure at point A when spigot C is closed?

Nesterboy [21]

Given,

$\begin{gathered} l_1=0.2m \\ l_2=0.07m \\ h=0.47m \\ \rho=\frac{1g}{cm^3} \end{gathered}$

Atm

3 0

2 years ago

A ball is revolving horizontally in a circle of radius 1.8 m, and is held by a rigid, massless rod. The mass of the ball is 0.1

OLga [1]

Answer:

$\omega=15 \ rad/s$

Explanation:

given,

radius of circle = 1.8 m

mass of the ball = 0.1 Kg

linear speed of the ball = 27 m/s

angular velocity of orbit = ?

$v = r \omega$

ω is the angular speed of the circle

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{27}{1.8}$

$\omega=15 \ rad/s$

the angular velocity of the orbit is $\omega=15 \ rad/s$

4 0

3 years ago

Does an object need a force to start moving

nadya68 [22]

Newton's first law actually states that a force isn't needed to keep an object moving, so no.

8 0

3 years ago