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Klio2033 [76]
3 years ago
6

Draw the Free- Body diagram of the 37 kilogram glass falling to the floor in a vacuum.

Physics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

362.6 N

Explanation:

F_{g}=mg

F_{g}=37\cdot9.8

F_{g}=362.6\ N

Therefore, the force that the glass hits the floor with is 362.6 N

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What is non uniform motion?​
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1 year ago
According to the "Law of Increasing Opportunity Costs," what would be the opportunity cost of a student who is staying up all ni
kodGreya [7K]

C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

<h3>What is law of opportunity cost?</h3>

The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.

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7 0
1 year ago
A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance
Minchanka [31]

Answer:

F=43570.9N

Explanation:

We can calculate the acceleration experimented by the passenger using the formula v_f^2=v_i^2+2ad, taking the initial direction of movement as the positive direction and considering it comes to a rest:

a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

F=ma=\frac{-mv_i^2}{2d}

Which for our values is:

F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N

6 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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