What will be the potential difference measured by an ideal voltmeter in the circuit of the figure?​

Answer:

(b)

Explanation:

The voltage always lags the current by 90°, regardless of the frequency.

Answer:

   a = w² r

Explanation:

In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.

           a = v² / r

angular and linear variables are related

           v = w r

we substitute

          a = w² r

where r is the radius of the wheel

Answer:

because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.

I see six (6) loops.

I attached a drawing to show where I get six loops from.

Answer:

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Explanation:

The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:

[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]

Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is

[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]

or

[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]

Answer:

3.05×10⁵ Nm²C⁻¹

Explanation:

According to Gauss' law,

∅' = q/e₀............... Equation 1

Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space

From the question,

Given: q = 2.7 μC = 2.7×10⁻⁶ C,

Constant: e₀ = 8.85×10⁻¹² C²/N.m²

Substituting these values into equation 1

∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)

∅' = 3.05×10⁵ Nm²C⁻¹

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

Answer:

The hot temperature is 157.5 K

The cold temperature is 48.8 K

Explanation:

Step 1: Data given

The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.

The volume increases by a factor of 5.7

The work output of the engine is 940 J in each cycle.

During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)

Step 2:For a carnot engine:

V2/V1 = V4/V3

Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))

⇒with Work = the work done in the cycle = 940J

⇒with n = the number of moles = 1.50 moles

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T1 = the hot temperature

⇒With T2⇒ the cold temperature

where R = 8.31 J/mol K Gas Constant

940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))

940 = 1.5 * 8.314 ln(2) * (T1-T2)

(T1-T2) = 940 / (1.5*8.314*ln(2))

(T1-T2) = 108.7K

For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)

For a monatomic ideal gas, Cv = 3/2R

When we combine both, we'll have:

T2 = T1*(1/5.7)^(R/3/2R)

T2 = T1*(1/5.7)^(2/3)

T2= T1 * 0.31

Since we know that (T1-T2) = 108.7K

we have:

T1 - 0.31T1= 108.7K

0.69T1 = 108.7K

T1 = 157.5K

T2 = 157.5*0.31 = 48.8K

The question is incomplete. The complete question is :

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].

Determine the following:

(a) frequency of the motion

(b) period of the motion

(c) amplitude of the motion

(d) first time after t = 0  that the object reaches the position x = 2.6 cm

Solution :

Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].

Comparing it with the general equation of simple harmonic motion,

 x = A sin (ωt + Φ)

  A = 4.7 cm

  ω = 7.9 π

a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]

                                             [tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]

                                             = 3.95 Hz

b). The period, [tex]$T=\frac{1}{f}$[/tex]

                        [tex]$T=\frac{1}{3.95}[/tex]

                            = 0.253 seconds

c). Amplitude is A = 4.7 cm

d). We have,

    x = A sin (ωt + Φ)

    [tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]

    [tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]

     [tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]

     [tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]

          Hence, t = 0.0236 seconds.

Answer:

v = 804.23 m/s

Explanation:

Given that,

The maximum distance covered by a cannon, d = 33 km = 33000 m

We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,

[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]

So, the initial velocity of the shell is 804.23 m/s.

Answer:

[tex]T=326.928K[/tex]

Explanation:

From the question we are told that:

Emissivity [tex]e=0.44[/tex]

Absorptivity [tex]\alpha =0.3[/tex]

Rate of solar Radiation [tex]R=0.3[/tex]

Generally the equation for Surface absorbed energy is mathematically given by

 [tex]E=\alpha R[/tex]

 [tex]E=0.3*950[/tex]

 [tex]E=285W/m^2[/tex]

Generally the equation for Emitted Radiation is mathematically given by

 [tex]\mu=e(\sigmaT^4)[/tex]

Where

T=Temperature

 [tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]

Therefore

 [tex]\alpha*E=e \sigma T^4[/tex]

 [tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]

 [tex]T=326.928K[/tex]

Answer:

Look at work

Explanation:

Elastic Collision: Ki=Kf

M1=4.65kg

M2: 0.060kg

v1=5m/s

v2=0m/s

4.65*5+0.060*0=4.65*v1'+0.060*v2'

23.25+0=4.65v1'+0.060v2'

Also since it is an elastic collision we can use

v1+v1'=v2+v2'

4.65+v1'=v2'

4.65+v1'=v2'

Substitute into the earlier equation

23.25=4.65v1'+0.060(4.65+v1')

Expand

23.25=4.65v1'+0.279+0.06v1'

Solve for v1'

22.971=4.71v1'

v1'=4.88m/s

v2'=4.65+4.88=9.53m/s

Answer:

E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules

Answer:

The answer is D. 2.25 × 1017 J

Explanation:

got it right on edge 2021

Answer:

It is a example of physical change

Answer:

7.84 m/s

Explanation:

Height, h = 2 m

Initial velocity, u = 10 m/s

Angle, A = 80°

(a) Let the time taken to go to the net is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]

Time cannot be negative.

So, t = 0.2 s

The vertical velocity at t = 0.2 s is

v = u + at

v = 10 sin 80 - 9.8 x0.2

v = 9.8 - 1.96 = 7.84 m/s

Answer:

The correct option is (E).

Explanation:

Given that,

Mass of object 1, m₁ = 1 kg

Mass of object 2, m₂ = 2 kg

They collides after the collision. We need to find the speed of the two boxes after the collision.

The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.

So, the information is not enough.

Answer:

9.89 m/s.

Explanation:

Given that,

The radius of the circular arc, r = 25 m

The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²

Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s[/tex]

So, the maximum speed of the car should be 9.89 m/s.

Answer:

W = Q * V     work done on charge Q

A. W = .5 C * 1.5 V = .75 Joules

B. P = W / t = .75 J / 1 sec = .75 Watts

Answer:

a) [tex] \alpha = 1.28 rad/s^{2} [/tex]  

b) Option ii. The angular acceleration would increase

Explanation:

a) The angular acceleration is given by:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular speed = v/r

v: is the tangential speed = 6.35 m/s

r: is the radius = 45.0 cm = 0.45 m

[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)

t: is the time = 11.0 s

α: is the angular acceleration

Hence, the angular acceleration is:

[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]  

b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).

Therefore, the correct option is ii. The angular acceleration would increase.

I hope it helps you!  

Answer:

Explanation:

Moment of inertia of solid sphere = 2/5 m R²

m is mass and R is radius of sphere.

Putting the values

Moment of inertia of solid sphere I₁

Moment of inertia of hollow  sphere I₂

Kinetic energy of solid sphere ( both linear and rotational )

= 1/2 ( m v² + I₁ ω²)                [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/5 m R² ω²)

= 1/2 ( m v² + 2/5 m v²)

=1/2 x 7 / 5 m v²

= 0.7 x 5 x 4² = 56 J .

This will be equal to work to be done to stop it.

Kinetic energy of hollow sphere ( both linear and rotational )

= 1/2 ( m v² + I₂ ω²)  [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/3 m R² ω²)

= 1/2 ( m v² + 2/3 m v²)

=1/2 x 5 / 3 m v²

= 0.833 x 5 x 4² = 66.64 J .

This will be equal to work to be done to stop it.

Answer:

SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.

Answer:

-700

formula is heat gained - work done

The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.

Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.

Given:

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,

Calculate the change in internal energy as shown below,

The change in internal energy = heat gained - work done

The change in internal energy = 1500 - 2200

The change in internal energy = -700 J

Thus, the change in internal energy is 700 joules.

To know more about Energy:

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The question is incomplete. The complete question is :

A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?

Solution :

Given :

Speed of the bullet train, v = 90 mi/h

                                            = [tex]$90 \times \frac{5280}{3600}$[/tex]

                                            = 132 ft/s

Time = 15 minutes

        = 15 x 60

        = 900 s

Acceleration from rest,

[tex]$a(t) = 4 \ ft/s^2$[/tex]

[tex]$v(t) = 4t + C$[/tex]

Since, v(0) = 0, then C = 0, so velocity is

v(t) = 4t ft/s

Then find the position function,

[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]

      [tex]$=2t^2+C$[/tex]

It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :

[tex]$s(t) = 2t^2$[/tex]

Time to get maximum cruising speed is :

4t = 132

t = 33 s

Distance travelled (at cruising speed) by speed to get the remaining distance travelled.

[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]

Total distance travelled, converting back to miles,

[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]

                      = 22.9125 mi

Therefore, the distance travelled is 22.9125 miles

The image is missing and so i have attached it.

Answer:

A) E = 8740 N/C

B) E = -6536 N/C

Explanation:

The formula for electric field is;

E = kq/r²

Where;

q is charge

k is a constant with value 8.99 x 10^(9) N•m²/C²

A) Now, to find the net electric field at point A, the formula would now be;

E = (kq1/(r1)²) - (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point A

r2 is distance from charge q2 to point A.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m

r2 = 10 cm = 0.1 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))

E = 8740 N/C

B) similarly, electric field at point B;

E = (kq1/(r1)²) + (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point B

r2 is distance from charge q2 to point B.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 10 cm = 0.1 m

r2 = 25cm + 10 cm = 35 cm = 0.35 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))

E = -6536 N/C

Answer:

B = 0.118 T

Explanation:

From Biot-Savart Law:

[tex]B = \frac{\mu_o I}{2\pi r}[/tex]

where,

B = strength of magnetic field = ?

μ₀ = 4π x 10⁻⁷ Tm/A

I = current enclosed = 474.1 A

r = radius = 0.8 mm = 8 x 10⁻⁴ m

Therefore,

[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]

B = 0.118 T

Answer:

0.000064 cubic meters.

Explanation:

Given the following data;

Length of side = 4 centimeters

Conversion:

100 centimeters = 1 meters

4 cm = 4/100 = 0.04 meters

To find the volume of cuboid;

Mathematically, the volume of a cuboid is given by the formula;

Volume of cuboid = length * width * height

However, when all the sides are equal the formula is;

Volume of cuboid = L³

Volume of cuboid = 0.04³

Volume of cuboid = 0.000064 cubic meters.

Answer:

810.94 m/s

Explanation:

Applying,

v = √(2gR)............. Equation 1

Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth

From the question,

Given: g = 0.636 m/s², R = 517 km = 517000 m

Substitute these values into equation 1

v = √(2×0.636×517000)

v = √(657624)

v = 810.94 m/s

Hence, the escape velocity is 810.94 m/s

Answer:

Look at work

Explanation:

Series:

I is the same for all resistors so just find the value of Req. In series Req= R1+R2+...+Rn. So here it will be 4.5+2.3=6.8ohms. Ieq=Veq/Req=4.41A. And since current is the same across all resistors the current to the lightbulb is 4.41A.

Parallel:

V is the same for all resistors so start of by finding Req. In parallel, Ieq=I1+I2+...+In. So I1= 30/4.5= 6.67A and I2= 13.04A. Ieq= 6.67+13.04= 19.71A.

Answer:

The correct answer is "1.2 J".

Explanation:

Seems that the given question is incomplete. Find the attachment of the complete query.

According to the question,

Now,

⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]

⇒      [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]

⇒      [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]

By putting the values, we get

⇒      [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]

⇒      [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]

⇒      [tex]=1.2 \ J[/tex]

Answer:

sup qwertyasdfghjk

Explanation: