During the 5 minute period described in question 4, the water in the insulated vessel undergoes a temperature increase of 20 C. Assuming all of the electrical energy dissipated by the resistor was transferred to the water as heat, what is the mass of the water

Answers

Answer 1

Answer:

Please find the complete question in the attached file.

Explanation:

[tex]V=12\ V\\\\I=1.2\ A\\\\T=5\times 60=300\ second\\\\[/tex]

Calculating the electrical energy dissipated:

[tex]w=p\cdot t=V\cdot I \cdot t\\\\[/tex]

              [tex]=12\times 1.2 \times 300 \ J\\\\=4320\ J[/tex]

[tex]\Delta T=20^{\circ}\ C\\\\W=m\cdot c\cdot \Delta T\\\\4320=m(4186 \times 20)\\\\m=\frac{4320}{4186 \times 20}=51.6 \ grams=0.516 \ kg\\\\[/tex]

During The 5 Minute Period Described In Question 4, The Water In The Insulated Vessel Undergoes A Temperature

Related Questions

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the x-axis does the third piece move

Answers

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical  component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

Answer:

8.3 m/s, 2196 degree from + X axis

Explanation:

m = 320 g , u = 2 m/s along X axis

m'  = 355 g, u' = 1.5 m/s along Y axis

m'' = 100 g, u'' = v

Let the speed of the third piece is v makes an angle A from the X axis.

use conservation of momentum along X axis

0 = 320 x 2 + 100 x v cos A

v cos A =  - 6.4 ..... (1)

Use conservation of momentum along Y axis

0 = 355 x 1.5 + 100 x v sin A

v sinA = - 5.3 ... (2)

Squaring and adding

[tex]v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s[/tex]

The angle is given by

[tex]tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree[/tex] from + X axis

gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be

Answers

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]

d = 68.5 x 10⁻⁶ m = 68.5 μm

what's impulse of
force

Answers

Answer:

The impulse experienced by the object equals the change in momentum of the object. In equation form, F.t = m. Δv. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.

A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.

1. How many revolutions do the kids make before the constant operational speed is reached ?

2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.

3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?

Answers

Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars

Answers

Answer:

Explanation:

I’ll help

An astronaut weighs 202 lb. What is his weight in newtons?

Answers

202 lb is 898.541newtons
The answer would be 898.541 newtons

Catching a wave, a 77-kg surfer starts with a speed of 1.3 m>s, drops through a height of 1.65 m, and ends with a speed of 8.2 m>s. How much nonconservative work was done on the surfer

Answers

Answer:

W = 2523.67 J

Explanation:

Given that,

The mass of surfer, m = 77 kg

He starts with a speed of 1.3 m/s

It drops through a height of 1.65 m and ends with a speed of 8.2 m/s.

We know that, the work done is equal to the change in kinetic energy. So,

[tex]W=\dfrac{1}{2}m(v_2^2-v_1^2)\\\\=\dfrac{1}{2}\times 77\times (8.2^2-1.3^2)\\W=2523.67\ J[/tex]

So, the required work done is equal to 2523.67 J.

15. Calculate the resistive forces acting on a sports car if it is travelling at a steady speed of 25 m/s when the engine is providing 200 kW. gnt V 10X50 - soot Grade 9



Answers

Answer:

8000 N

Explanation:

Applying

P = F×V.............. Equation 1

Where P = Power of the sport car, F = resistive force acting on the sport car, V  = Speed of the sport car.

make F the subject of the equation

F = P/V............ Equation 2

From the question,

Given: P = 200 kW = 200000 W, V = 25 m/s

Substitute these values into equation 2

F = 200000/25

F = 8000 N

Work-Energy Theorem & Power
A 0.5 kg mass sitting on smooth ice is accelerated from rest by a force until is
acquires a speed of 8 m/s. The force acts while the mass moves through a
displacement of 2 m.
A. Calculate the kinetic energy of the mass after the force acts.
B. Calculate the work done by the force.
C. Calculate the magnitude of the force that accelerated the mass.

Answers

Answer:

A. 16 J

B. 16 J

C. 8 N

Explanation:

A. Determination of the kinetic energy.

Mass (m) = 0.5 Kg

Velocity (v) =. 8 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 8²

KE = ½ × 0.5 × 64

KE = 0.5 × 32

KE = 16 J

B. Determination of the Workdone by the force.

Kinetic energy (KE) = 16 J

Workdone =.?

Workdone and kinetic energy has the same unit of measurement. Thus,

Workdone = kinetic energy

Workdone = 16 J

C. Determination of the force.

Workdone (Wd) = 16 J

Displacement (s) = 2 m

Force (F) =?

Wd = F × s

16 = F × 2

Divide both side by 2

F = 16 / 2

F = 8 N

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop

Answers

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

[tex]V_i=50 \ m/h[/tex]

Now,

[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]

    [tex]=73.333 \ ft/sec[/tex]

Now,

Final velocity,

[tex]V_f=0[/tex]

Initial velocity,

[tex]V_{initial} = 73.333 \ ft/sec[/tex]

hence,

⇒ [tex]V_f^2=V_i^2+2aD[/tex]

By putting the values, we get

      [tex]0=(73.333)^2+2\times( -22) D[/tex]

  [tex]44D=(73.333)^2[/tex]

      [tex]D=\frac{(73.333)^2}{44}[/tex]

          [tex]=122.2211[/tex]

An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16

Answers

Answer:

answer : option (b) 0.016 amp

explanation : resistance of resistor , R = 10 Ω

inductance of inductor , X_LX

L

= 20H

voltage of AC circuit , V = 120volts

frequency, ff =60Hz

so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s

now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=

R

2

2

L

2

V

= 120/√{10² + (120π)² × 20²}

= 120/√{100 + 14400π² × 400}

after solving this we get, i = 0.016 amp

A police car travels towards a stationary observer at a speed of 15m/s. the siren on the car emits a sound of frequency 250Hz. Calculate the observer frequency. the speed of sound is 340m/s​

Answers

Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)

= 250 x (340/( 340-15))

= 261.54 Hz

The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns

A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.

Answers

Answer:

C

Explanation:

Similarity

describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please

Answers

Answer:

Force A newton Law first law

F = M.A which Force in 0 N as you Questions Above

Force B

Newton Law 3

Action = -Reaction

Hope you can explain this formula as you want to scribe to explaining

1. A child slide down an inclined plane of length 10 m at an angle of 45°. If the coefficient friction between the child and the plane is 0.1, evaluate The velocity just before touching the bottom of the plane.

Answers

Answer:

The speed at the bottom is 11.2 m/s.

Explanation:

length, s = 10 m

Angle, A = 45 degree

coefficient of friction = 0.1

let the velocity is v.

The acceleration is given by

[tex]a = g sin A - \mu g cos A \\\\a = 9.8 (sin 45 - 0.1 cos 45)\\\\a = 6.24 m/s^2[/tex]

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 6.24 \times 10 \\\\v = 11.2 m/s[/tex]

The upward normal force exerted by the floor is 710 N on an elevator passenger who weighs 720 N . You may want to review (Pages 107 - 110) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Weighing yourself in an elevator. Part A What is the reaction force to the upward normal force exerted by the floor

Answers

Answer:

If the person is to remain the floor the reaction force will be equal to the normal force exerted by the floor.

F(normal) - F(reaction) = 0

That means the person is not moving with respect to the elevator.

Expanding the applied forces we have:

Fw - Fn = 720 - 710 = 10 N   where the positive direction is chosen as down

Fw is the weight of the person and Fn the force exerted on the person by the elevator,

The acceleration of the person the becomes F = m a = m * 10 N and will be downward agreeing with our choice of coordinate axes.

Which car has the greatest potential energy? What type of potential energy does the
car have? What characteristics or properties does potential energy depend on? Which
cars have the least potential energy? Explain.

Answers

Explanation:

*car A has the greatest PE

*gravitational potential energy

*mass and height

*car D has the least PE

During the Moment of Inertia experiment, a group of students decided to mount a solid sphere on the top of the horizontal disk. The sphere and disk have the same mass, M, and radius, R. They recorded the values for the initial and final positions for the falling mass, and the height above the floor when falling mass is at its lowest position. These values are respectively 90cm, 68.5 cm, and 23.5cm. The radius of the pulley, r, was 1/10 of the radius of the sphere. Furthermore, the mass of the disk, M, was 1.4kg, what was the mass, m, for the falling mass

Answers

Answer:

u are the very good person I know if u will do itself u will becam3 a rising star

Explain why it takes much more effort to stop a freight train compared with a car?

Answers

Answer:

Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.

Question 15
Calculate the velocity of a body if its total energy is three times its rest energy
OA 0.54c
OB. 0.760
OC0.94c
OD.C
A Moving to another question will save this response.
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Type here to search
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BI
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Answers

Jjdjdjfjififififkjfndndj

What process provides the sun with its energy

Answers

Answer:

nuclear fusion

The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom. During the fusion process, radiant energy is released.

Answer:

nuclear fusion

Explanation:

The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.

The reason why a teacher is more important then a farmer

Answers

Answer:

A teacher is more important than a famer.

Explanation:

A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.

Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến
95oC for 20 minutes.

Answers

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

[tex]Power = P = \frac{E}{t}[/tex]

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

[tex]P = \frac{2.927\ x\ 10^6\ J}{1200\ s}[/tex]

P = 2439.5 W = 2.439 KW

A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with the axis. On the left and right faces, the field has a strength and , respectively. The field along the front and back faces has strengths and . The field at the bottom and top faces has strengths and , respectively. What is the total charge enclosed by the cube

Answers

Complete Question

Complete Question is attached below

Answer:

[tex]q=1.558*10^{-9}c[/tex]

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength [tex]E_l=784.75N/m[/tex]

Right field strength [tex]E_r=776.38 N/m[/tex]

Front field strength [tex]E_f=725.5 N/m[/tex]

Back field strength [tex]E_b=749.54 N/m[/tex]

Top field strength [tex]E_t=944.95 N/m[/tex]

Bottom field strength [tex]E_{bo}=1082.58 N/m[/tex]

Generally, the equation for  Charge flux is mathematically given by

[tex]\phi=EAcos\theta[/tex]

Where

Theta for Right,Left,Front and Back are at an angle 90

[tex]cos 90=0[/tex]

Therefore

[tex]\phi =0[/tex] with respect to Right,Left,Front and Back

Generally, the equation for  Charge Flux is mathematically also given by

[tex]\phi=\frac{q}{e_o}[/tex]

Where

[tex]Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2[/tex]

Therefore

[tex]Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t[/tex]

[tex]Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)[/tex]

[tex]Q_{net}=176N/C m^2[/tex]

Giving

[tex]q=\phi*e_0[/tex]

[tex]q=176N/C m^2*1.558*10^{-12}c[/tex]

[tex]q=1.558*10^{-9}c[/tex]

Physics question on picture

Answers

Answer:

B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction

The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.

Answers

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         [tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex]            (you has an mistake in the formula)

         [tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]

         [tex]\frac{1}{C_{eq1}}[/tex] = 0.1   10⁶

         [tex]C_{eq1}[/tex] = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          [tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]

          [tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]

          [tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶

          [tex]C_{eq2}[/tex] = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

           

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]

          U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]

          U₄ = 0.5 J

To get maximum current in a circuit, the resistance should be in _____
1)series
2)parallel

Answers

Answer:

no parallel is the correct answer

What is the electric field strength between two parallel conducting plates separated by 10 cm and having a potential difference between them of 2000 V?

a.
2000 V/m

b.
200 V/m

c.
20 kV/m

d.
200000 V/m

Answers

Answer:

• Potential Difference (V) = 2000 V

• Distance b/w the two parallel plates (d) = 10 cm = 10/100 = 1/10 = 0.1 m

• Electric field (E) = ?

[tex]\implies V = E.d[/tex]

[tex]\implies E = \dfrac{V}{d} [/tex]

[tex]\implies E = \dfrac{2000}{0.1} [/tex]

[tex]\implies E = \dfrac{2000}{ {10}^{ - 1} } [/tex]

[tex]\implies E = 2000 \times {10}^{1} [/tex]

[tex]\implies\bf E = 20000 \:V/m[/tex]

[tex]\implies\bf E = 20\:kV/m[/tex]

Hence, option C) the correct answer.

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.

Required:
a. What is the net force on the side of the container?
b. How long does it take and how much the water level will drop till water no longer comes out of the faucet?

Answers

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = [tex]\frac{P_o ( 1-0.850)}{\rho \ g}[/tex]

        h = [tex]\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}[/tex]

        h = 1.55 m

A vehicle is used to transport material down a straight aisle. The max acceleration of the vehicle is 1 m/s/s and the max speed of the vehicle is 5m/s. The vehicle starts at the beginning of the aisle. How long does it take to move down the aisle and come to a stop at the other end if:
a) the aisle is 100 meters long?
b) the aisle is 9 meters long?

Answers

Answer:

(a) 14.14 s

(b) 4.24 s

Explanation:

maximum acceleration, a =  1 m/s^2

maximum speed, v = 5 m/s

initial speed, u = 0 m/s

(a) distance, s = 100 m

Let the time is t.

Use second equation of motion

[tex]s = u t 0.5 at^2\\\\100 = 0 + 0.5 \times 1 \times t^2\\\\t = 14.14 s[/tex]

(b) distance, s = 9 m

Let the time is t'.

Use second equation of motion

[tex]s = u t + 0.5 at^2\\\\9= 0 + 0.5 \times 1 \times t'^2\\\\t' = 4.24 s[/tex]

Other Questions
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