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4 years ago

A penny dropped into a wishing well reaches the bottom in 2.50 seconds. What was the velocity at impact?

Physics

2 answers:

evablogger [386]4 years ago

7 0

Does anyone wanna premarital seggs?

12345 [234]4 years ago

4 0

Answer:

Taking it was dropped with inital velocity of 0m/s.

v = u+at

v = 0 + 9.8*1.5

v = 14.7s

(a=9.8m/s^2 << That's the acceleration due to gravity)

Explanation:

You might be interested in

Does sound travels through your skull to your ear when you speak ? If yes then how ?

Maksim231197 [3]

Sound travels through waves, more specifically, through vibrations. They do not go from skull to ear, but they can go from ear to brain, or skull to brain. Ear to brain is simply vibrations traveling from outer ear, to inner ear, to the brain. Skull to brain, otherwise known as "bone conduction", has the vibrations hitting the skull, then to the temproal bone, then to the inner ear where the brain picks it up.

5 0

3 years ago

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

If the acceleration of an object is zero at some instant in time, what can be said about its velocity at that time? 1. It is neg

Mkey [24]

3. It is not changing at that time

Explanation:

If the acceleration of a body is zero at some instant in time, it implies that the velocity is not changing at that point in time. Velocity is the rate of change of displacement with time.

✓Acceleration and velocity shares a very close relationship.

✓ For a body to accelerate, the velocity must change. Acceleration is defined as the rate of change of velocity with time.

✓If at any point, a body moves with constant velocity i.e the velocity does not change with time, the acceleration becomes zero.

✓ For acceleration to occur, a body must change velocity.

Learn more:

Acceleration brainly.com/question/6323625

#learnwithBrainly

5 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

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

If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1,000 km

astra-53 [7]

Answer:

change 1000km into Metre

1000km=1000000 M

speed= distance/time

time=distance/speed

time=1000000/299792458

time= 0.0033second

6 0

3 years ago