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3 years ago

if the mass of a material is 50 grams and the volume of the material is 15 cm^3 what would the density of the material be?

Physics

1 answer:

Lostsunrise [7]3 years ago

7 0

density=mass/volumr ... 50/15 gm/cc ... 3.3333333333

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When a physical change in a sample occurs, which of the following would NOT change?

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Which simple machine is described as an inclined plane wrapped around a cylinder?

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Calculate the amount of heat needed to increase the temperature of 200 grams of water at 10°C to 95°C.

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the amount of heat need is 71060 Joules

Explanation

The equation for the amount of heat needed to increase the temperature of water is as follows:

$\begin{gathered} E=mC(T_2-T_1) \\ where \\ \\ \end{gathered}$

where E is the amount of heat

C is the specific heat capacity of water having a standard value of 4180 J/kg*K

T is the final ans initial temperature in kelvin

m is the mass

Step 1

given

$\begin{gathered} mass=m=200\text{ gr=0.2 kg} \\ T_1=10\text{ \degree C} \\ T_2=95\text{ \degree C} \\ C=4180\text{ J/kh*K} \end{gathered}$

a) convert the temperature into kelvin

$\begin{gathered} 10\text{ \degree C=\lparen10+273\rparen K=283 K} \\ 95\text{ \degree C=\lparen95+273 \rparen K=368 K} \end{gathered}$

now, replace in the formula

$\begin{gathered} E=mC(T_{2}-T_{1}) \\ E=0.2\text{ kg*4180L/kg*K*\lparen368K-283 k\rparen} \\ E=71060\text{ J} \end{gathered}$

therefore, the amount of heat need is 71060 Joules

I hope this helps you

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1 year ago

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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