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2 years ago

One of the Lady Spartans was falling to the ground after

Physics

1 answer:

dem82 [27]2 years ago

0 0

Answer:

Explanation:

Given that,

A lady falling has a final velocity of 4m/s

v = 4m/s

Mass of the lady is 60kg.

m = 60kg

Using conservation of energy, the potential energy of the body from the point where the lady is dropping is converted to the final kinetic energy of the lady.

Therefore,

P.E = K.E(final) = ½mv²

P.E = ½ × 60 × 4²

P.E = 480 J.

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A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

BaLLatris [955]

The answer would be 2.8m height on earth takes
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>

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2 years ago

Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?

Ksenya-84 [330]

Answer:

The answer is " $6\ \frac{m}{s}$ "

Explanation:

The formula for velocity:

$\to \overline{v}={\frac{\Delta x}{\Delta t}}$

$=\frac{6}{1}\\\\=6\ \frac{m}{s}$

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1 year ago

When the voltage and current have _____ polarities in a pure capacitive circuit, the capacitor is discharging and the energy is

tangare [24]

Answer:opposite

Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed

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1 year ago

A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t

Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature $T_1=323K$

Volume $V_1=V\ and\ V_2=\frac{V}{2}$

As there is no heat transfer so this is an adiabatic process

For and adiabatic process $TV^{\gamma -1}=constant$

Here $\gamma =1.4$

So $T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}$

$T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1$

$T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K$

4 0

2 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

2 years ago