PHYSICS COLLEGE

Which is the SI base unit for mass?

Answers

Answer 1

Answer:

Answer:

kilogram

Explanation:

Answer 2

Answer:

Answer:

SI base units of mass=KG

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HELP ME PLS!!!!Find the location of beryllium (Be) on the periodic table. What type of ion willberyllium form?A. An ion with a -2 chargeB. An ion with a +6 chargeC. An ion with a +2 chargeD. An ion with a -6 charge

As you take the stoppered part of the tube up the staircase you begin to see the water level drop around the 4th floor. As you continue up it does not continue up with you but stays at a constant level. What does that mean?a. The pressure in the tubing is equal to the barometric pressure.
b. The tubing was unable to supply any more water to the tube for use.
c. The pressure outside the tube is higher that the water pressure inside the tube.

Answers

Answer:

a. The pressure in the tubing is equal to the barometric pressure.

Explanation:

Since in the question it is mentioned that the if you take the stoppert part of the tube than the level of warer would be fall approx 4th floor and if it is continued than it wont be continue but remains constant.

Now here first we do that the tube i.e. connected to the bucket should be taken up. In the first instance, the bucket supplies the water to the tube but it would not increased far away to the level of the barometric pressure

Hence, the correct option is a.

An AC voltage source has an output of ∆V = 160.0 sin(495t) Volts. Calculate the RMS voltage. Tries 0/20 What is the frequency of the source? Tries 0/20 Calculate the voltage at time t = 1/106 s. Tries 0/20 Calculate the maximum current in the circuit when the generator is connected to an R = 53.8 Ω resistor.

Answers

Answer:

RMS voltage is 113.1370 V

frequency is 780.685 Hz

voltage is −158.66942 V

maximum current is 2.9739 A

Explanation:

Given data

∆V = 160.0 sin(495t) Volts

so Vmax = 160

and angular frequency = 495

time t = 1/106 s

resistor R = 53.8 Ω

to find out

RMS voltage and frequency of the source and voltage and maximum current

solution

we know voltage equation = Vmax sin ωt

here Vmax is 160 as given equation in question

so RMS will be Vmax / √2

RMS voltage = 160/ √2

RMS voltage is 113.1370 V

and frequency = angular frequency / 2π

so frequency = 497 / 2π

frequency is 780.685 Hz

voltage at time (1/106) s

V(t) = 160.0 sin(495/ 108)

voltage = −158.66942 V

so current from ohm law at resistor R 53.8 Ω

maximum current = voltage max / resistor

maximum current = 160 / 53.8

maximum current = 2.9739 A

Final answer:

The root-mean-square voltage of the AC source is 113.14 V, its frequency is 78.75 Hz, and the voltage at time t = 1/106 s is approximately 150.4 V. The current at this peak voltage, when connected to a resistor of 53.8 Ω, is approximately 2.97 A.

Explanation:

The output of an AC voltage source can be represented by the equation V = V₀ sin ωt, where V₀ is the peak voltage, ω is the angular frequency, and t is the time. In this case, V₀ = 160 V and ω = 495 (1/s). The root-mean-square voltage (Vrms), which is commonly used to express AC voltage, can be calculated from the peak voltage using the formula Vrms = V₀/√2 which gives approximately 113.14 V.

The frequency of the source is related to the angular frequency by the equation f = ω/2π, which gives a frequency of approximately 78.75 Hz. To find the voltage at a specific time t = 1/106 s, we substitute these values into the initial equation resulting in V = V₀ sin ωt = approximately 150.4 V.

Finally, the resistance R = 53.8 Ω allows us to calculate the maximum current in the circuit given by I = V/R. The maximum current occurs at the peak voltage, so I(max) = V₀/R = approximately 2.97 A.

Learn more about AC Voltage Source here:

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Light from a sodium vapor lamp (λ-589 nm) forms an interference pattern on a screen 0.91 m from a pair of slits in a double-slit experiment. The bright fringes near the center of the pattern are 0.19 cm apart. Determine the separation between the slits. Assume the small-angle approximation is valid here.

Answers

Answer:

separation between the slits is 0.28 mm

Explanation:

given data

wave length λ = 589 nm = 589 × $10^(-9)$ m

distance between slits and the screen D = 0.91 m

fringes weight y = 0.19 cm = 0.19 × $10^(-2)$ m

solution

we find here the spacing between the two slits i.e d

so use here formula that is

y = λD ÷ d .........................1

put here value we get

0.19 × $10^(-2)$ = $(589*10^(-9)*0.91)/(d)$

solve we get

d = 0.28 mm

A 55-liter tank is full and contains 40kg of fuel. Find using Sl units: • Density p. • Specific Weight y • Specific Gravity Answer tolerance = 1%. Be sure to include units. The sign of the answers will not be graded, use a positive value for your answer. Your answers: p= (Enter a positive value) y = (Enter a positive value) SG = (Enter a positive value)

Answers

Answer:

Density of the fuel is 727.3 kilograms per cubic meter.

Specific weight of the fuel is 7127.3 Newtons per cubic meter.

Specific gravity of the fuel is 0,727.

Explanation:

In order to use SI units, we have to convert liters to cubic meters. Knowing that a liter is a cubic decimeter and a cubic decimeter is $1*10^(-3)$ cubic meters, we know that the tank has 0,055 cubic meters of fuel (because it is full).

Now that we have things in SI units, we calculate density:

$p_(fuel)= (mass)/(volume) = (40 kg)/(0.055 m^(3) ) =727.3 (kg )/(m^(3) )$

Knowing the mass per unit of volume, we can calculate weight per unit of volume thanks to Newton's second law (mass times acceleration, g in this case, equals force (weight)), i.e. specific weight:

$y=p*g=727,3 (kg)/(m^(3))*9.8(m )/(s^(2))=7127,3 (N)/(m^(3))$

With density we can also calculate how dense the fuel is related to a reference (water), i.e. specific gravity. SG is a dimensionless number that tell us how much denser (SG>1) or lighter per unit of volume (SG<1) a substance is than water. We use water as a reference because it is one of the most used substances in our life, and it is a standard density (1000 kg per cubic meter at 4°C and 1 atm).

$SG=(p_(fuel) )/(p_(water) ) =(727.3 (kg )/(m^(3) ))/(1000 (kg )/(m^(3) )) =0,727$

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answers

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

The additional resistance is $R_z = 4.4 \Omega$

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = (E)/(R_T)$

Substituting values

$4 = (28)/(R_z + 2.6)$

Making $R_z$ the subject of the formula

So $R_z = (28 - 10.4)/(4)$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

A pilot performs an evasive maneuver by diving vertically at 270. If he can withstand an acceleration of 9.0 's without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea? y= ?m