An electrical appliance draws 9.0 amperes of current when connected to a 120-volt
source of potential difference. What is the total amount of power dissipated by this
appliance?
1100 W
110 W
13 W
130 W

Answer:

In which school you are???

Explanation:

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

Answer: 0.00693

Explanation:

Given

Spring constant [tex]k=2\ N/m[/tex]

Mass of object [tex]m=75\ g[/tex]

The amplitude of the oscillation decreases from 10 mm to 5 mm  in 15 s

Equation of amplitude for the ideal spring-mass system is

[tex]\Rightarrow A=A_oe^{-\frac{bt}{2m}}\quad \quad [\text{b=damping constant}]\\\text{Insert the values}\\\\\Rightarrow 5=10e^{\frac{b\times 15}{2\times 0.075}}\\\\\Rightarrow e^{-\frac{b\times 15}{2\times 0.075}}=0.5\\\\\text{Taking natural log both sides}\\\\\Rightarrow \ln \left(e^{-\frac{b\times 15}{2\times 0.075}}\right)=\ln 0.5\\\\\Rightarrow -\dfrac{15b}{0.15}=-0.693\\\\\Rightarrow b=0.00693[/tex]

Answer:

Cold

Explanation:

Im pretty sure im sorry if I am wrong

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

So they give us this

V=IR

V= 1.8

I=0.4

R=?

So we insert the thing that we know.

1.8=0.4*R

We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.

So we have this.

Lastly we solve.

R=4.5ohms

The formula to find R is V=IR

V/I=R

So the resistance will be the Voltage divided by the Current

Answer:

The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

Explanation:

Given;

radius of the circular loop, r = 2.0 m

maximum induced emf, E = 5.0 V

The emf induced in a magnetic field is given as;

[tex]emf = \frac{d\phi}{dt} \\\\\phi = AB\\\\emf = A\frac{dB}{dt} \\\\\frac{dB}{dt} = \frac{emf}{A} \\\\where;\\A \ is \ the \ area \ circular \ l00p = \pi r^2 = \pi (2)^2 = 4\pi \ m^2\\\\\frac{dB}{dt} = \frac{5}{4\pi} \\\\\frac{dB}{dt} = 0.398 \ T/s[/tex]

Therefore, the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

Answer:

The right solution is "0.511".

Explanation:

Given:

Initial moment of inertia,

= 1630 kg.m²

Radius,

= 5 m

Angular speed,

= 1.6 rad/s

Now,

The moment of inertia after stepping on will be:

= [tex]1630+2\times (69.5\times (5)^2)[/tex]

= [tex]1630+2\times (69.5\times 25)[/tex]

= [tex]5105 \ Kg.m^2[/tex]

hence,

As per the question, the angular speed is conserved, then

⇒ [tex]1630\times 1.6=5105\times \omega'[/tex]

            [tex]2608=5105\times \omega'[/tex]

                [tex]\omega'=\frac{2608}{5105}[/tex]

                     [tex]=0.511[/tex]

Answer:

Explanation:

comfortable distance is 25 cm .

He must be using convex lens . In that case rays coming from object placed at 25 cm appears to be coming from 59.1 cm due to converging nature of convex lens.

object distance u = -25 cm

image distance v = -59.1 cm

Lens formula

1 / v - 1 /u = 1 /f

-1 / 59.1 + 1 / 25 = 1/f

- .0169 + .04 = 1 / f

.0231 = 1 / f

f = 43.3 m

Answer:

  α = 0.357 ras / s²

Explanation:

This is a rotational kinematics exercise, it tells us that it performs 10 squats in 30 s, for which it performs one squat at t = 3 s, also indicates that the angle of the squat is θ = 92º

            θ = θ₀ + w₀ t + ½ α t²

the athlete starts from rest, whereby w₀ = 0 and the initial angle in the vertical position is zero (θ₀=0)

            θ = ½ α t²

            α = 2 θ /t²

let's reduce the magnitudes to the SI system

            θ = 92º (π rad /180º) = 0.511π rad

let's calculate

            α = 2 0.5111π /3²

            α = 0.1136π rad / s²

            α = 0.357 ras / s²

Answer:

377 nm

Explanation:

Number of lines per meter is, [tex]N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}[/tex]

Grating element is, [tex]d=\frac{1}{N}[/tex]

[tex]=1.8868 \times 10^{-6} \mathrm{~m}[tex]

Order is, n=5

Condition for maximum intensity is, [tex]d \sin \theta=n \lambda[/tex]

 [tex]\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\ &=0.377 \times 10^{-6} \mathrm{~m} \\ &=377 \mathrm{~nm}[/tex]

Answer:

To protect us.

Explanation:

For ex. your dunking on a basketball hoop if that wasn't strong you would fall on your back and get injured.

Answer:

The person's velocity is zero.

Explanation:

Answer:

a)   I₁ / I₂ = 9,  b)    I₁ / I₂ = 25

Explanation:

a) The amount of light that can pass through a telescope is the intensity of the light

         I = P / A

where A is the area of ​​the mirror and P is the incident power that is the same in all telescopes

       P = I₁ A₁ = I₂ A₂

where subscript 1 is for the 2 m diameter telescope and subscript 2 for the 6 m diameter telescope

The mirror area is

        A = π r²

         r = d / 2

we substitute

        I₁ π d₁² /4 = I₂ π d₂² ​​/ 4

        I₁ / I₂ = (d₂ / d₁) ²

we calculate

        I₁ / I₂ = (6/2) ²

        I₁ / I₂ = 9

b) d₂ ​​= 15 inch and d₁ = 3 inc

let's calculate

          I₁ / I₂ = (15/3) ²

          I₁ / I₂ = 25

The sum of the resistance = 2 ohms x 80 lights = 160 ohms.

Current = Total voltage / total resistance:

Current = 120V / 160 ohms

Current = 0.75 Amps

Answer:

Explanation:

Rate of volume flow at two points will be same at two points .

A₁ V₁ = A₂V₂

A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .

A₁ = 18.2 x 3.55 = 64.61 m²

V₁ = 2 .55 x 10⁻² m/s

A₂ = 16.3 x d = 16.3 d  m²

d is depth at second point .

V₂ = 11.6 x 10⁻² m/s

64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d  m² x 11.6 x 10⁻² m/s

d = .87 m

so canal is .87 m deep.

Answer:

a. hydroelectric power plant

Answer:

♕ [tex]\large{ \red{ \tt{Step - By - Step \: Explanation}}}[/tex]

☃ [tex] \underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} : [/tex]

♨ [tex] \underline {\underline{ \orange{ \large{ \tt{T \: O \: \: F \: I \: N\: D}}}} }: [/tex]

☄ [tex]\underline{ \underline{ \large{ \pink{ \tt{S\: O \: L \: U \: T\: I \: O \: N}}}}}: [/tex]

✧ [tex] \red{ \boxed{ \large{ \purple{ \sf{Wave \: velocity(v) = Frequency(f) \times Wavelength(λ)}}}}}[/tex]

~Plug the known values and then multiply!

↦ [tex] \large{ \tt{7 \times 42}}[/tex]

↦ [tex] \boxed{ \boxed{ \large{ \bold{ \tt{294 \: m {s}^ {- 1} }}}}}[/tex]

☥ [tex] \large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}} [/tex]

---------------------------------------------------------------

❁ [tex] \underline{ \large{ \red{ \tt{D\: E\: T \: A \: I \: L\: E \: D \: \: I\: N \: F \: O}}}} : [/tex]

# KILL : Excuses

KISS : Opportunities

MARRY : Goals

♪ Hope I helped! ♡

☂ Have a wonderful day / night ! ツ

✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✔

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

Answer:

Explanation:

b) light

The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is

[tex]W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J[/tex]

Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.

The only other force acting on the block as it moves is the force P. Let [tex]W_P[/tex] be the work done by the force P. Then the total work done on the block is

[tex]W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}[/tex]

Answer:

That is equal to R1 + R2. If three or more unequal (or equal) resistors are connected in series then the equivalent resistance is: R1 + R2 + R3 +…, etc. One important point to remember about resistors in series networks to check that your maths is correct.

Answer:

T = 1.12 10⁻⁷ s

Explanation:

This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.

All the charge dq is at a distance r

           dE = k dq / r²

Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry

            cos θ =[tex]\frac{dE_x}{dE}[/tex]

             dEₓ = dE cos θ

              cos θ = x / r

substituting

                dEₓ = [tex]k \frac{dq}{r^2 } \ \frac{x}{r}[/tex]

                DEₓ = k dq x / r³

let's use the Pythagorean theorem to find the distance r

             r² = x² + a²

where a is the radius of the ring

we substitute

              dEₓ = [tex]k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq[/tex]

we integrate

               ∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} }  ∫ dq

               Eₓ = [tex]k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}[/tex]

In the exercise indicate that the electron is very central to the center of the ring

                x << a

                Eₓ = [tex]k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}[/tex]

if we expand in a series

                  [tex](\ 1+ (x/a)^2 \ )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2[/tex]

we keep the first term if x<<a

                 Eₓ = [tex]\frac{ k Q}{a^3} \ x[/tex]

the force is

                 F = q E

                 F = [tex]- \frac{kQ }{a^3} \ x[/tex]

this is a restoring force proportional to the displacement so the movement is simple harmonic,

                 F = m a

                 [tex]- \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }[/tex]

                 [tex]\frac{d^2 x}{dt^2} = \frac{keQ}{ma^3} \ x[/tex]

the solution is of type

                  x = A cos (wt + Ф)

with angular velocity

                w² = [tex]\frac{keQ}{m a^3}[/tex]

angular velocity and period are related

                 w = 2π/ T

we substitute

               4π² / T² = \frac{keQ}{m a^3}

                T = 2π  [tex]\sqrt{\frac{m a^3 }{keQ} }[/tex]

let's calculate

                T = 2π [tex]\sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19} \ 0.021 \ 10^{-3} } }[/tex]

                 T = 2π pi [tex]\sqrt{320.426 \ 10^{-18} }[/tex]

                 T = 2π  17.9 10⁻⁹ s

                 T = 1.12 10⁻⁷ s

Answer:

The air in the soil is similar in composition to that in the atmosphere with the exception of oxygen, carbon dioxide, and water vapor. In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.036%.

hope this helps

have a good day :)

Explanation:

Answer: how the government and teachers can address or help students experiencing stress and anxiety.

Explanation: