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RideAnS [48]
3 years ago
14

You are riding on a school bus and suddenly get thrown forward. What did the bus just do?

Physics
1 answer:
sergij07 [2.7K]3 years ago
5 0
The bus is going forward and suddenly stops
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A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at
Naddika [18.5K]

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

4 0
3 years ago
Using the right amount of significant figures, calculate the answer to the following problem, 215.5+101.02555
cestrela7 [59]

Answer:

This is how I figured it out:

  1. 215.5 rounded to one significant figure is 200
  2. 101.02555 rounded to one significant figure is 100.
  3. 200 + 100 = 300.

Hope this helps!

Explanation:

7 0
3 years ago
What renewable source of energy would be suited to chicago IL?
Kay [80]

Answer:

Wind is the primary renewable resource used for electric power generation in the state. In 2019, wind provided 97% of the state's renewable energy generation, and Illinois was sixth in the nation in utility-scale (1 megawatt or greater) wind capacity, with about 5,200 megawatts online.

Explanation:

7 0
2 years ago
What word means what scientists do that is the basis for their investigation
Mademuasel [1]
Scientific Method: Because, they use their method which the scientific one to solve investigations.

7 0
3 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
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