A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

[tex]Q_c = 110 \ J[/tex]

Coefficient of performance refrigerator,

[tex]Cop(refrig)=4[/tex]

(a)

As we know,

⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]

or,

⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]

              [tex]=\frac{110}{4}[/tex]

              [tex]=27.5 \ Joules[/tex]

(b)

⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]

or,

⇒ [tex]Q_h = Q_c+Work[/tex]

         [tex]=114+27.5[/tex]

         [tex]=141.5 \ Joules[/tex]

Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

We know that from Ohm's law

V= IR

Now  R= V/I

Substitute

R= 0.032/1.35

R= 0.023 Ohms

Hence the resistance is 0.023 Ohms

Answer:

g = 0.0114 m/s²

Explanation:

The value of acceleration due to gravity on the surface of the moon can be given by the following formula:

[tex]g = \frac{Gm}{r^2}[/tex]

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m = mass of moon = 4.2 x 10¹⁶ kg

r = radius of moon = 1.57 x 10⁴ m

Therefore,

[tex]g= \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(4.2\ x\ 10^{16}\ kg)}{(1.57\ x\ 10^4\ m)^2}[/tex]

g = 0.0114 m/s²

Answer:

Explanation:capacitor

Answer:

Ammeter

Explanation:

pls mark me as a brainlist

Answer:

x = A sin (wt + theta)        where w = angular frequency - basic SHM equation

w = 8 pi = 2 pi f

f = 4         basic frequency

N = 8     number of times thru origin

Each cycle the particle will pass thru the origin +x and -x    twice

Answer:

prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)

please sove step by step with language it is opt maths question

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

Determine the mass transfer coefficient ( m/s )

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

Answer:

Work done = 2289.084 Joules

Explanation:

Given the following data;

Mass = 102 Kg

Height = 2.29

Time = 1.32 seconds

We know that acceleration due to gravity, g = 9.8 m/s²

a. To find the work done by the person;

Here, work would be done in the form of gravitational potential energy.

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

Work done = 102 * 2.29 * 9.8

Work done = 2289.084 Joules

Explanation:

Given:

[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]

[tex]V = 12 V[/tex]

[tex]I = 1.2 A[/tex]

Recall that power P is given by

[tex]P = VI[/tex]

so the amount of energy dissipated [tex]\Delta E[/tex] is given by

[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]

[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

Answer:

The technician will need to prepare 48 syringes for a 24 hour supply since the patient needs double the available dose.

Explanation:

brainliest plz . . .

Answer:

I THINK C

Explanation:

BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.

Explanation:

Given that,

In order to find the distance travelled, firstly we need calculate the acceleration.

→ v = u + at

→ 20 = 0 + 60a

→ 20 = 60a

→ 20 ÷ 60 = a

→ ⅓ m/s² = a

Now, by using the 2nd equation of motion :

→ s = ut + ½at²

→ s = 0(60) + ½ × ⅓ × (60)²

→ s = ⅙ × 3600

→ s = 1 × 600

→ s = 600 m

Hence, the distance travelled is 600 m.

Answer:

2.9::: 5.87*10*3 N/C

8: 7.73 × 10 ^2  N/C

Explanation: https://study.com/academy/answer/a-solid-metal-sphere-of-radius-3-00-m-carries-a-total-charge-of-5-50-muc-what-is-the-magnitude-of-the-electric-field-at-each-of-the-following-distances-from-the-sphere-s-center-a-3-10-m-b-8-00-m.html

Answer: no sorry../

Explanation:

Answer:

velocity of ship wrt sea= 7i∧

velocity of women on deck= 40j∧

velocity of women relative to sea will be resultant of above two velocoties,

7i∧ + 40j∧ magnitude is

square root (7 x 7 + 40x40)

=√49+1600

=√1612

=40.14 m/s

Answer:

B) the medium

Explanation:

B. Depends on the medium.

Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

Answer:

[tex]D_l=d[/tex]

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

[tex]V_{e1}=V_{e2}[/tex]

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

[tex]D=\frac{at^2}{2}[/tex]

Where

Acceleration is given as

[tex]a=\frac{V_o}{2d}[/tex]

And

Time

[tex]T=\frac{d}{v_0}[/tex]

Therefore horizontal displacement towards the left is

[tex]D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}[/tex]

[tex]D_l=d[/tex]

Answer:

To cover a larger distance we use km(kilometre) , hm (hectometre) , and dac(decametre). These are called the multiples of Metre. For the distance is smaller , we use Dm (decimetre , cm(centimetre) and mm (millimetre) . These are called the submultiples of

Answer:

explained

Explanation:

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.

A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.

Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.

Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at  5  m/s

straight toward the goal and drives the ball in the same direction with a velocity of  30 m/s

relative to her body, then the velocity of the ball is  35  m/s

relative to the stationary, profusely sweating goalkeeper standing in front of the goal.

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (

The figure shows components of velocity v in horizontal  vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.

Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors  and  

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.

Answer:

always runs slower than normal.

Explanation:

The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.

According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.

In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.

Answer: 26.84 m/s

Explanation:

Given

Original frequency of the horn [tex]f_o=228\ Hz[/tex]

Apparent frequency [tex]f'=246\ Hz[/tex]

Speed of sound is [tex]V=340\ m/s[/tex]

Doppler frequency is

[tex]\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)[/tex]

Where,

[tex]v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}[/tex]

Insert values

[tex]\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s[/tex]

Thus, the speed of the car is [tex]26.84\ m/s[/tex]

Answer:

a)   θ = 14.23º, b)   θ₂ = 75.77,  c) t = 0.6019 s

Explanation:

This is a missile throwing exercise.

a) the reach of the ball is the distance traveled for the same departure height

          R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]

          sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]

          sin 2θ = 7.00 9.8 / 12.0²

          2θ = sin⁻¹ (0.476389) = 28.45º

           θ = 14.23º

the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º

          θ ’= 90  -14.23

          θ’= 75.77º

b) the two angles that give the same range are

         θ₁ = 14.23

         θ₂ = 75.77

the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.

C) the time of the pass can be calculated with the expression

           x = v₀ₓ t

           t = x / v₀ₓ

           t = 7 / 11.63

           t = 0.6019 s

Answer:

v = speed

c = speed of light

using the equation

L = Lc Sqrt( 1 - (v/c)2)

9 = 14 Sqrt( 1 - (v/c)2)

v/c = 0.765

v = 0.765 c

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

Answer:

Mass is zero if the whole rope is horizontal. Imaginary rope.

Explanation:

With any mass at all, only a small section of the rope will be truly horizontal. The rope curve will be a a catenary.