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3 years ago

How long will it take for a rock to fall 50 meters

Physics

1 answer:

denpristay [2]3 years ago

7 0

Answer:

The force of gravity, g = 9.8 m/s2

Gravity accelerates you at 9.8 meters per second per second. After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.

Time to splat: sqrt ( 2 * height / 9.8 )

It's the square root because you fall faster the longer you fall.

The more interesting question is why it's times two: If you accelerate for 1 second, your average speed over that time is increased by only 9.8 / 2 m/s.

Velocity at splat time: sqrt( 2 * g * height )

This is why falling from a higher height hurts more.

Energy at splat time: 1/2 * mass * velocity2 = mass * g * height

Explanation:

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Andy has set a goal to be able to run 65 laps on the PACER by the end of the school year. This is an example of a ____.

lana66690 [7]

Answer:

The answer is long- term goal

Explanation:

6 0

3 years ago

An electric motor has an effective resistance of 22 22 and an inductive reactance of 72 12 when working under load. The voltage

Alecsey [184]

Answer:

Current amplitude, I = 5.57 A

Explanation:

It is given that,

Effective resistance, R = 22 ohms

Inductive reactance, L = 72 ohms

Voltage of the alternating source, V = 420 volt

The total impedance of the RL circuit is given by :

$Z=\sqrt{R^2+L^2}$

$Z=\sqrt{(22)^2+(72)^2}$

Z = 75.28 ohms

From Ohm's law,

$V=I\times Z$

$I=\dfrac{V}{Z}$

$I=\dfrac{420}{75.28}$

I = 5.57 A

So, the current amplitude of the electric motor is 5.57 A. Hence, this is the required solution.

6 0

4 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0

3 years ago

A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through

aniked [119]

Answer:

$a_2=24.5\ \text{m/s}^2$ towards right

$a_1=2.1875\ \text{m/s}^2$ towards left

Explanation:

$m_1$ = Mass of skater = 56 kg

$m_2$ = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2$

So, the ball will move towards right with a magnitude of acceleration $a_2=24.5\ \text{m/s}^2$ .

The force on the ball will be

$F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}$

The force on the ball is $122.5\ \text{N}$

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

$-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2$

So, the skater will move towards left with a magnitude of acceleration $2.1875\ \text{m/s}^2$

4 0

3 years ago

A 0.3-m-radius automobile tire rotates how many revolutions after starting from rest and accelerating at a constant 2.13 rad/s2

Aloiza [94]

Answer:

The automobile tire rotates 91 revolutions

Explanation:

Given;

angular acceleration of the automobile, α = 2.13 rad/s²

time interval, t = 23.2-s

To calculate the number of revolutions, we apply the first kinematic equation;

$\theta = \omega_i \ + \frac{1}{2} \alpha t^2$

the initial angular velocity is zero,

$\theta =0\ + \frac{1}{2} (2.13) (23.2)^2\\\\\theta = 573.2256 \ Rad$

Find how many revolutions that are in 573.2256 Rad

$N = \frac{\theta}{2 \pi} = \frac{573.2256}{2\pi} \\\\N = 91 \ revolutions$

Therefore, the automobile tire rotates 91 revolutions

3 0

4 years ago