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Juliette [100K]
3 years ago
15

If a roller coaster cart needs more Kinetic Energy to reach the top of the track, what are two ways to increase the amount of Ki

netic Energy the cart has? Provide evidence to support your answer.
Physics
1 answer:
Neko [114]3 years ago
3 0

Answer:

Increase the mass or velocity.

Explanation:

The formula of Kinetic Energy is Ek=1/2*m*v^2. So both increase the mass or velocity can increase the Ek.

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A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage o
aleksley [76]

Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

We know that from Ohm's law

V= IR

Now  R= V/I

Substitute

R= 0.032/1.35

R= 0.023 Ohms

Hence the resistance is 0.023 Ohms

5 0
3 years ago
100 joule of heat produced each 1 secomd on 4ohm find the potential difference ​
irakobra [83]

Answer:

20 V

Explanation:

Power is 100 J/s or 100 W.

We know that P = IV = \frac{V^{2} }{R} .

Isolate the potential difference. V = \sqrt{RP} = \sqrt{100 * 4} = 20 V

8 0
3 years ago
What causes seasons to occur?
LUCKY_DIMON [66]
It’s C I hope it helps you
3 0
3 years ago
Read 2 more answers
A system consists of electrons and protons only. It contains 150 electrons and has a total charge of +22e. What is the mass of t
liubo4ka [24]

According to the statements the number of electrons is 150, then

e = 150

But there is a positive charge of +22e, then the number of protons would be

p = 150+172

If the mass of the electrons is

m_e = 9.11*10^{-31} kg

And the mass of the protong is

m_p = 1.673*10^{-27}kg

We have that the total mass of the system would be

m = e*m_e +pm_p

m = 150 * (9.11*10^{-31})+170(1.673*10^{-27})

m = 2.84547*10^{-25} kg

5 0
3 years ago
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
3 years ago
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