Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb

Answers

Answer 1

Answer:

i₀ = V / R_i

Explanation:

For this exercise we use Ohm's law

         V = i R

          i = V / R

the equivalent resistance for

         [tex]\frac{1}{R_{eq}}[/tex] =  ∑ [tex]\frac{1}{R_i}[/tex]

if all the bulbs have the same resistance, there are N bulbs

         [tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]

         R_{eq} = R_i / N

we substitute

         i = N V / Ri

where i is the total current that passes through the parallel, the current in a branch is

         i₀ = i / N

         i₀ = V / R_i


Related Questions

A TV satellite dish is designed to receive radio waves of wavelength
0.0644 meters. What is the frequency of the waves it receives? _______GHz

Give your answer in gigahertz (GHz). 1 GHz = 10^9 Hz.

Give your answer to the nearest tenth of a GHz (one place after the decimal). Just enter the number; do NOT use scientific notation.

Answers

Answer:

4.7 GHz

Explanation:

Applying,

v = λf................. Equation 1

Where v = velocity of the radio wave, λ = wavelength, f = frequency

make f the subject of the equation

f = v/λ.............. Equation 2

Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s

From the question,

Given: λ = 0.0644 meters

Constant: v = 3.00 x 10⁸ m/s

Substitute these values into equation 2

f = (3.00 x 10⁸)/0.0644

f = 4.66×10⁹ Hz

f = 4.7 GHz

Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volume and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB

Answers

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

[tex]V_A = \frac{1}{2}V_B[/tex]

Now, from Charle's Law:

[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]

Hence, the correct option is:

A. TA = TB/2.

Explain the following observations:
a) A balloon filled with hydrogen gas floats in air;
B) A ship made of steel floats on water.

Answers

Answer and Explanation:

a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.

b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.




A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[

Answers

the thermometer is the temperature that is around it so its registering the temperature its supposed to

if the tin is made of a metal which has a density of 7800 kg per metre cubic calculate the volume of the metal used to make tin and lead​

Answers

Answer:

XL sleep usual Addison officer at home and ear is not a short time to be be free and ear is a short time to make a short time

Explanation:

so that I can take the class on Monday and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to time for a day or night and ear buds is Anshu and duster and duster fgor a day or night is Anshu and duster for a day or not a week of computer science from your computer and I am in the same as I am a short of ti and you can be the first time I will be be

Why is the force of attraction between the Earth and ourselves so huge compared to the attraction between two apples?

Answers

Answer:

Answer in explanation

Explanation:

The force of attraction between two bodies is governed by Newton's Law of Gravitation:

[tex]F = \frac{Gm_1m_2}{r^2}[/tex]

where,

G = Universal Gravitational Constant

m₁ = mass of the first body

m₂ = mass of the second body

r = distance between the two bodies

F = Force

Hence, it is clear from the formula that the magnitude of the force is directly proportional to the product of the masses of the objects. So in the case of the earth and ourselves, the mass of the earth is very large in order of 10²⁴ kg. Due to this huge mass, the attraction between the earth and ourselves is so huge as compared to the attraction between two apples. Because the masses of the apple are very small in grams.

Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then the power transferred to the turbine will be: a) 100 KW b) 10 KW c) 1 KW d) 100 W​

Answers

Answer:

Natae Si Jordan Kaya Sya Napaihe

Explanation:

haha

A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
a. Plot the data as engineering stress versus engineering strain.
b. Compute the modulus of elasticity.
c. Determine the yield strength at a strain offset of 0.002.
d. Determine the tensile strength of this alloy.
e. What is the approximate ductility, in percent elongation?
f. Compute the strain energy density up to yielding (modulus of resilience).
( Load in N Load in lb Length in mm Length in in. 2.000 2.002 2.004 2.006 2.008 2.010 2.020 2.040 2.080 2.120 2.160 2.200 2.240 2.270 2.300 2.330 Fracture 50.800 7330 15,100 3400 23,100 5200 30,400 6850 34,400 7750 38,400 8650 41,3009300 44,800 10,100 46,200 10,400 53, 47,300 10,650 54.864 47,500 10,700 55.880 46,100 10,400 44,800 10,100 42,600 9600 3,400 8200 Fracture Fracture Fracture 50.851 50.902 50.952 51.003 51.054 1650 51.308 51.816 52.832 848 56.896 57.658 58.420 59.182

Answers

Answer:

A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).

Topic: Chapter 10: Projectory or trajectile?
Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?

90°
75°
45
50°
30°

Answers

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

A new car manufacturer advertises that their car can go from zero to sixty mph in 8 [s]. This is a description of

Answers

Answer:

Acceleration

Explanation:

The fact that new can go from zero to 60mph in 8 secs is a description of its pick-up or in physics,  it's called acceleration.

Here initial velocity u= 0

final velocity v = 60 mph = 1m/minute.

or v =1609.344/60 = 26.82m/s

and time taken to do so is 8 sec

Acceleration a = (v-u)/t

a = (26.82-0)/8 = 3.35 m/s^2

Therefore, acceleration of the car a = 3.35 m/s^2.

As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.

Answers

Answer:

frequency

Explanation:

The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.

So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is  

[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]

where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.

A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press
5.0 m from the right support column (Figure slide 8). Calculate the force
on each of the vertical support columns.

Answers

Answer:

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

[tex]\sum F = 0[/tex] and [tex]\sum \tau = 0[/tex]

From the image, expressing the forces through the y-axis, we have:

[tex]F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}[/tex]

Also, let the force [tex]F_1[/tex] be the pivot and computing the torque to determine [tex]F_2[/tex]:

Then:

[tex]F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P[/tex]

[tex]F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}[/tex]

[tex]F_2 = 117600 \ N[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

For the force equation:

[tex]F_1+F_2=1.617*10^5 \ N;[/tex]

where:

[tex]F_2 = 1.176*10^5 \ N[/tex]

Then:

[tex]F_1+1.176*10^5 \ N=1.617*10^5 \ N[/tex]

[tex]F_1=1.617*10^5 \ N-1.176*10^5 \ N[/tex]

[tex]F_1=44100\ N[/tex]

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC​

Answers

Explanation:

Given:

L = 0.02 H

C = [tex]2\:\mu \text{F}[/tex]

f = 200 Hz

The general form of the impedance Z is given by

[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to

[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]

[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]

The velocity-time graph of a body is given. What quantities are represented by (a) slope of the graph and (b) area under the graph?​

Answers

Answer:

a) acceleration

b) displacement

Explanation:

The velocity-time graph is a graph of velocity versus time. The velocity (m/s) would be on the Y-axis while time (s) would be on the X-axis.

a) The slope of a graph is given by: change in Y-axis/change in X-axis = ΔY/ΔX

In a velocity-time graph, ΔY = change in velocity and ΔX = change in time.

Hence, the slope of a velocity-time graph becomes: change in velocity/change in time.

Also, acceleration = change in velocity/change in time.

Hence, the slope of a velocity-time graph = acceleration.

b) Assuming that the area under a velocity-time graph is a rectangle, the area is given as:

Area of a rectangle = length x breadth

                                  = velocity x time (m/s x s)

Also, displacement = velocity x time (m)

Hence, the area under a velocity-time graph of a body would give the displacement of the body.

A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)​

Answers

Answer:

K = 0 J

Explanation:

Given that,

The mass of the particle, m = 1.2 mg

The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]

We need to find the kinetic energy of the particle at time t = 0 s.

At t = 0 s, the particle is at rest, v = 0

So,

[tex]K=\dfrac{1}{2}mv^2[/tex]

If v = 0,

[tex]K=0\ J[/tex]

So, the kinetic energy of the particle at time t = 0 s is 0 J.

One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×10^3 m/s, given the collision lasts 6.00×10^8s.

Answers

Answer:

F = 6666.7 N

Explanation:

Given that,

Mass of a chip, m = 0.1 mg

Initial speed, u = 0

Final speed,[tex]v=4\times 10^{3}\ m/s[/tex]

Time of collision,[tex]t=6\times 10^{-8}\ s[/tex]

We know that,

Force, F = ma

Put all the values,

[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N[/tex]

So, the required force is 6666.7 N.

PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?

Answers

[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]

Explanation:

Given:

[tex]\lambda = 4.96×10^3 s[/tex]

[tex]A_0 = 3.21x10^{17}[/tex] nuclei

t = 1.98×10^4 s

[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]

[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]

[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]

At which point is the kinetic energy of the pendulum the greatest?

C

A

D

B

Answers

Answer:

Point C

Explanation:

Greatest Kinetic Energy means lowest potential energy since energy is conserved. Lowest potential energy means lowest height which is at Point C.

The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?​

Answers

Answer:

7.9 [tex]\frac{m}{s^{2} }[/tex]

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]

Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:

[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]

Plug everything in:

7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]

(1590kg the initial weight plus the weight of the added passenger)

A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p

Answers

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere=[tex]\rho[/tex]

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

[tex]Density=\frac{mass}{volume}[/tex]

[tex]Mass=Density\times volume=Constant[/tex]

Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]

Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]

Using the formula

[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]

[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]

[tex]\rho_2=8\rho[/tex]

Hence, the density of  the compressed sphere=[tex]8\rho[/tex]

Option a is correct.

Preocupada com o aumento da tarifa na conta de luz, uma pessoa resolve economizar diminuindo o tempo de banho de 20 para 15 minutos. Seu chuveiro possui as seguintes especificações: 4200 W e 220V. Sabendo que o kWh custa R$0,30, a economia feita em 10 dias foi de aproximadamente​

Answers

Answer:

The mount saved is $ 0.105.

Explanation:

Concerned about the increase in the electricity bill, a person decides to save by reducing bathing time from 20 to 15 minutes. Your shower has the following specifications: 4200 W and 220V. Knowing that the kWh costs R$0.30, the savings made in 10 days were approximately​.

The electrical energy is given by

E = P x t

where, P is the electrical power and t is the time.

When he is using the shower for 20 minutes, the energy consumed is

E = 4200 x 20 x 60 = 5040,000 J = 1.4 kWh

When he is using the shower for 15 minutes, the energy consumed is

E' = 4200 x 15 x 60 = 3780000 J = 1.05 kWh

The difference in energy is

E'' = E - E' = 1.4 - 1.05 = 0.35 kWh

The money saved is

= 4 0.3 x 0.35 = $ 0.105

When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.

Answers

Answer:

angular acceleration.

Explanation:

Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.

Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.

A 1.64 kg mass on a spring oscillates horizontal frictionless surface. The motion of the mass is described by the equation: X = 0.33cos(3.17t). In the equation, x is measured in meters and t in seconds. What is the maximum energy stored in the spring during an oscillation?

Answers

Answer:

[tex]K.E_{max}=0.8973J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.64kg[/tex]

Equation of Mass

[tex]X=0.33cos(3.17t)[/tex]...1

Generally equation for distance X is

[tex]X=Acos(\omega t)[/tex]...2

Therefore comparing equation

Angular Velocity [tex]\omega=3.17rad/s[/tex]

Amplitude A=0.33

Generally the equation for Max speed is mathematically given by

[tex]V_{max}=A\omega[/tex]

[tex]V_{max}=0.33*3.17[/tex]

[tex]V_{max}=1.0461m/s[/tex]

Therefore

[tex]K.E_{max}=0.5mv^2[/tex]

[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]

[tex]K.E_{max}=0.8973J[/tex]

Two children sit on a seesaw that is in rotational equilibrium. The first child has weight W and sits at distance d from the pivot. If the second child sits at a distance of 7*d from the pivot, what must be the weight of the second child

Answers

Answer:

W/7

Explanation:

By principle of moments,

Sum of clockwise moment = sum of anticlockwise moment

Weight × 7d = W × d

Weight = W/7

Since the two children are in rotational equilibrium, the weight of the second child is W/7.

How can the weight of the second child be determined?

The weight of the second child can be determined from the principle of moments.

The principle of moments states that for a body in equilibrium, the sum of the clockwise moments and anticlockwise moments about a point is zero.

Let the weight of the second child be X

From the principle of moments:

W × d = 7×d × X

X = W/7

Therefore, the weight of the second child is W/7.

Learn more about principle of moments at: https://brainly.com/question/20298772

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration

Answers

Answer:

a= -80.357 m/s

Explanation:

use the formula

vf^2=vi^2+2a(xf-xi)

Plug in givens

0=(7.50)^2+2a(0.350m)

solve for acceleration

a= -80.357 m/s

how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?​

Answers

Assuming no heat lost to the surrounding,

-5⁰C ice → 0⁰C ice

Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(2.0 x 10³) x (0-(-5))

Q = 50000J

0⁰C ice → 0⁰C water

Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg

Q = mLf

Q = 5(3.34 x 10⁵)

Q = 1670000J

0⁰C water → 100⁰C water

Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(4.2 x 10³) x (100-0)

Q = 2100000J

100⁰C water → 100⁰C steam

Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Q = mLv

Q = 5(2.26 x 10⁶)

Q = 11300000J

Total amount of heat required

= 50000 + 1670000 + 2100000 + 11300000

= 15120000J

Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2

Answers

Solution :

Part A .

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.

So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km

           [tex]y[/tex] component is = 4 x cos (29°) = 3.498 km

Part B

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]

So the [tex]x[/tex] component is = -2 cm/s

           [tex]y[/tex] component is = 0

Part C

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.

So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]

           [tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]

The x- and y-components of the vectors  is mathematically given as as follows for each Part respectively

x= -1.939 km, y= 3.498 km

x= -2 cm/s, 0

y=, x= -7.6412m/s^2, -10.517m/s^2

What are the x- and y-components of the vectors?

Question Parameters:

Generally, we follow a basic principle where

x component= Fsin\theta

y component= Fcos\theta

Therefore

For A

x component is

x= -4 x sin (29°)

x= -1.939 km

 y component is

y= 4 x cos (29°)

y= 3.498 km

For B

x component is

x= -2 cm/s            

y component is

y= 0

For C

x component is

x= -13 x sin (36°)

x= -7.6412m/s^2      

y component is

y= -13 x cos (36°)

y= -10.517m/s^2  

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A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?

Answers

Answer:

7.8% of the original volume.

Explanation:

From the given information:

Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C

Pressure [tex]P_1[/tex] = 240 kPa

Temperature [tex]T_2[/tex] = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

[tex]P_1V_1 =nRT_1[/tex]

making V_1 (initial volume) the subject:

[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]

[tex]V_1 = \dfrac{nR*295}{240}[/tex]

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume [tex]V_2[/tex] can be computed as:

[tex]V_2 = \dfrac{nR*318}{240}[/tex]

Now, the change in the volume ΔV =  V₂ - V₁

[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]

[tex]\Delta V = \dfrac{23nR}{240}[/tex]

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]

[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]

[tex]= 0.078[/tex]

= 7.8% of the original volume.

suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along

Answers

Answer:

Pressure is more in the open container than the closed one.

Explanation:

The pressure due to the fluid at a depth is given by

Pressure = depth x density of fluid x gravity

So, when the container is open, the atmospheric pressure is also add  up but when  the container is closed only the pressure due to the fluid is there.

So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.

hen the container is closed only the pressure due to the fluid is there.

Question 3 of 10
Which statement describes the law of conservation of energy?
A. Air resistance has no effect on the energy of a system.
B. Energy cannot be created or destroyed.
C. The total energy in a system can only increase.
D. Energy cannot change forms.
هما
SUBMIT

Answers

Answer:

B . energy cannot be created or destroyed

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