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3 years ago

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A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How far did the bulle

t travel horizontally when it hit the ground?
595.2 m
478.4 m
364.0 m
247.2 m

2 answers:

Vesnalui [34]3 years ago

5 0

It’s most likely A , i’m not so sure though

USPshnik [31]3 years ago

4 0

Answer:

im pretty sure its a

Explanation:

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What is the gpe of a 200 kg hot air ballon 21,000 m above the ground?

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here we have

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$GPE = 200(9.8)(21000)$

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Solve this chemical equation: CH3CH2OH+__O2=CO2+__H2O

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3 years ago

A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu

seropon [69]

Answer:

$\theta=50\ revolution$

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, $\omega_i=600\ rpm=62.83\ rad/s$

Finally it shuts off, $\omega_f=0$

Time taken, t = 10 s

Let $\alpha$ is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{0-62.83}{10}$

$\alpha =-6.283\ rad/s^2$

Let $\theta$ is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }$

$\theta=\dfrac{-62.83^2}{2\times -6.283}$

$\theta=314.15\ radian$

$\theta=49.99\ revolution$

or

$\theta=50\ revolution$

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0

3 years ago

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

b)

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