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4 years ago

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A 50kg hiker is standing on the edge of a cliff. Find the hiker’s gravitational potential energy if the cliff is 30m high. (Show

the equation, show your work and answer with units.)

2 answers:

DENIUS [597]4 years ago

4 0

Answer:

The answer to your question is: Ep = 14715 Joules

Explanation:

Data

mass = 50 kg

Potential energy = ?

height = 30 m

g = 9.81 m/s²

Equation

Ep = mgh

Substitution

Ep = (50)(9.81)(30)

Ep = 14715 Joules

CaHeK987 [17]4 years ago

3 0

Answer:

14715 joules

Explanation:

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A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

6 0

3 years ago

How much energy does the electron have initially in the n=4 excited state?what is the change in energy if the electron from part

Ierofanga [76]

What it looks to be that you found in A was the "initial"...b/c the question asks:
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>

<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>

The answer comes to: <span>−1.36×10^−19 J</span>

You use the same equation for the second part as for part a.
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>

8 0

3 years ago

Marvin uses a long copper wire with resistivity 1.68 x 10^-8 Ω⋅m and diameter 1.00 x 10^−3 m to create a solenoid that has a 3

Mazyrski [523]

Answer with Explanation:

We are given that

Resistivity of copper wire= $\rh0=1.68\times 10^{-8}\Omega m$

Diameter=d= $1.00\times 10^{-3} m$

Radius of copper wire= $r=\frac{d}{2}=\frac{1}{2}\times 10^{-3} m$

Radius of solenoid=r' $=3 cm=3\times 10^{-2} m$

1 m=100 cm

a.Length of wire=l=11.3 m

Area of wire=A= $\pi r^2$

Where $\pi=3.14$

A= $3.14\times (\frac{1}{2}\times 10^{-3})^2$

Resistance, R= $\rho \frac{l}{A}$

Using the formula

$R=1.68\times 10^{-8}\times\frac{11.3}{3.14\times (\frac{1}{2}\times 10^{-3})^2}$

$R=0.24\Omega$

B.Length of solenoid= $2\pi r'=2\times 3.14\times 3\times 10^{-2}=0.188$ m

Number of turns= $n_0=\frac{l}{2\pi r'}=\frac{11.3}{0.188}$

$n_0$ =60

C.Potential difference,V=3 V

Current,I= $\frac{V}{R}$

I= $\frac{3}{0.24}=12.5 A$

D.Total length =0.1 m

Number of turns per unit length,n= $\frac{60}{0.1}=600$

Magnetic field along central axis inside of the solenoid,B= $\mu_0 nI$

$B=4\pi\times 10^{-7}\times 12.5\times 600=9.42\times 10^{-3} T$

4 0

3 years ago

Read 2 more answers

After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that

almond37 [142]

After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

<h3>What is Half-Life?</h3>

Half-Life refers to the time it takes for half the amount of a substance to disappear or change.

The nucleus of the atoms of radioactive elements disintegrate to half their starting amounts after every Half-Life.

After three half-lives one-eight of the original atoms remain.

Therefore, after three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

Learn more about Half-Life at: brainly.com/question/26689704

#SPJ1

7 0

2 years ago

A copper cylinder has a mass of 76.8 g and a specific heat of 0.092 cal/g.C. It is heated to 86.5 degrees C from 19.5 degrees C.

Sladkaya [172]

Given:

The mass of the copper cylinder is: m = 76.8 g = 0.0768 kg

The change in the temperature is: T = 86.5 deg C - 19.5 deg C = 67 deg C

The specific heat is: c = 0.092 cal/g.C

To find:

Heat energy needed to heat the copper cylinder.

Explanation:

The specific heat is defined as the amount of heat energy needed to raise the temperature of a substance by one degree celcius.

The expression relating heat Q, mass m, specific heat c and temperature difference T is:

$Q=mcT$

Substitute the values in the above equation, we get:

$\begin{gathered} Q=76.8\text{ g}\times0.092\text{ Cal/g.C}\times67\text{ deg C} \\ \\ Q=473.40\text{ Cal} \end{gathered}$

Final answer:

473.40 calories of heat is required to heat the copper cylinder.

6 0

1 year ago