A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.

Answer:

KE_2 = 3.48J

Explanation:

Conservation of Energy

E_1 = E_2

PE_1+KE_1 = PE_2+KE_2

m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²

(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2

4.10J+0.914J = 1.53J + KE_2

5.01J = 1.53J + KE_2

KE_2 = 3.48J

Answer:

[tex]T=1.54s[/tex]

Explanation:

From the question we are told that:

Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]

Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]

Therefore

Frequency of Motor 1 [tex]f_1=13.75[/tex]

Frequency of Motor 2  [tex]f_2= 13.1[/tex]

Generally the equation for Time Elapsed is mathematically given by

[tex]T=\frac{1}{df}[/tex]

Where

[tex]df=f_1-f_2[/tex]

[tex]df=13.75-13.1[/tex]

[tex]df=0.65Hz[/tex]

Therefore

[tex]T=\frac{1}{65}[/tex]

[tex]T=1.54s[/tex]

Answer:

a)  [tex]v=206.896m/s[/tex]

b)  [tex]L=6.749m[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=46Hz[/tex]

Length [tex]l=579m[/tex]

Total Mass [tex]T=4.3kg[/tex]

Tension [tex]T=3423[/tex]

a)

Generally the equation for velocity is mathematically given by

[tex]v=\sqrt{\frac{T}{\rho}}[/tex]

Where

[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]

Therefore

[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]

[tex]v=206.896m/s[/tex]

b)

Generally the equation for length of string is mathematically given by

[tex]L=\frac{3\lambda}{2}[/tex]

Where

[tex]\lambda=\frac{v}{f}[/tex]

[tex]\lambda=\frac{206.89}{46}[/tex]

[tex]\lambda=4.498[/tex]

Therefore

[tex]L=\frac{3*4.498}{2}[/tex]

[tex]L=6.749m[/tex]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

[tex]L = \frac{N^2 \mu_0 A}{l}[/tex]

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

[tex]A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2[/tex]

[tex]L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH[/tex]

(b) The energy stored in the inductor when 21 A current ;

[tex]E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J[/tex]

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

[tex]emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s[/tex]

The car has initial speed 50 mi/h ≈ 73 ft/s, so it covers a distance x such that

0² - (73 ft/s)² = 2 (-24 ft/s²) x

==>   x ≈ 111.0 ft

Answer:

2.3  ×  [tex]10^{-1}[/tex]  

Explanation:

1 kg = 1000 g.

0.00023 kg x 1000 g = 0.23 grams

Answer:

0.23×10⁴

Explanation:

kilogram to gram ÷ 1000

0.00023kg ÷ 1000

=0.23g

scientific notation=0.23×10⁴

The aspect of the US justice system that has its roots in Jewish scripture is:

the idea that all people are subject to the same rules and laws.

It is the doctrine of "equality before the law."  Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc.  Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism.  It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.

Answer:

answer is C

the idea that all people are subject to the same rules and laws

Explanation:

hope this helps!

(1) When the driver is at rest, the restoring force exerted by spring is equal in magnitude to the driver's weight, so that

∑ F = s - mg = 0   ==>   s = mg = 617.4 N

If the spring is compressed 0.015 m, then the spring constant k is such that

617.4 N = k (0.015 m)   ==>   k = 41,160 N/m ≈ 41 kN/m

(2) The total mass of the passengers is

24 (68 kg) + 3 (15 kg) + 5 (3 kg) + 25 kg = 1717 kg

so that if everyone is at rest, the spring is compressed a distance x such that

kx = (1717 kg) g   ==>   x ≈ 0.41 m

Answer:

The mass of Al is 89.027 kilograms.

The mass of Jo is 104.973 kilograms.

The magnitude of the force of Jo on Al is 596.481 newtons.

Explanation:

Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:

Al:

[tex]m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t[/tex] (1)

Jo:

[tex]m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t[/tex] (2)

Total mass:

[tex]m_{1} + m_{2} = 194\,kg[/tex]

Where:

[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the skaters, in kilograms.

[tex]v_{1,o}[/tex], [tex]v_{1,f}[/tex] - Initial and final velocities of Al, in meters per second.

[tex]v_{2,o}[/tex], [tex]v_{2,f}[/tex] - Initial and final velocities of Jo, in meters per second.

[tex]F[/tex] - Impact force between skaters, in newtons.

[tex]\Delta t[/tex] - Impact time, in seconds.

If we know that [tex]v_{1,o} = 0\,\frac{m}{s}[/tex], [tex]v_{1,f} = -7.9\,\frac{m}{s}[/tex], [tex]\Delta t = 1\,s[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex] and [tex]v_{2,f} = 6.7\,\frac{m}{s}[/tex], then the masses of the skaters are, respectively:

[tex](194-m_{2})\cdot (-7.9) = -F[/tex] (1b)

[tex]m_{2} \cdot 6.7 = F[/tex] (2b)

(2b) in (1b):

[tex](194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7[/tex]

[tex]-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}[/tex]

[tex]14.6\cdot m_{2} = 1532.6[/tex]

[tex]m_{2} = 104.973\,kg[/tex]

[tex]m_{1} = 194\,kg - 104.973\,kg[/tex]

[tex]m_{1} = 89.027\,kg[/tex]

And the magnitude of the force is:

[tex]F = 6.7\cdot m_{2}[/tex]

[tex]F = 596.481\,N[/tex]

The mass of Al is 89.027 kilograms.

The mass of Jo is 104.973 kilograms.

The magnitude of the force of Jo on Al is 596.481 newtons.

Answer:

in which topic you need help

the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).

Explanation:

please mark me as a brainlieast

This question is incomplete, the complete question is;

A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.

(a) Determine the magnitude of the emf induced in the loop.

(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)

Answer:

a) the magnitude of the emf induced in the loop is 0.003 V

b) dA/dt = 0.002 m²/s

Explanation:

Area of the loop wire A = 0.015 m²

magnitude of the field is increasing dB/dt = 0.20 T/s

a)

Determine the magnitude of the emf induced in the loop.

V = A( dB/dt )

we substitute

V = 0.015 m² × 0.20 T/s

V = 0.003 V

Therefore,  the magnitude of the emf induced in the loop is 0.003 V

b)  the induced emf is;

V = B( dA/dt ) + A( dB/dt )

given that; induced emf is 0, B = 1.5

so we substitute

0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]

-[ 1.5T × ( dA/dt )] = 0.003 m²T/s

dA/dt = -[ 0.003 m²T/s / 1.5T ]

dA/dt = -0.002 m²/s

the negative shows that the area is decreasing

hence, dA/dt = 0.002 m²/s

Answer:

the required angle is 0.0834879⁰

Explanation:

Given  the data in the question;

slit separation; d = 1.75 mm = 1.75 × 10⁻³ m

wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m

wavelength λ₂ 510 nm = 510 × 10⁻⁹ m

Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;

tanθ = [tex]y_m[/tex] / D = mλ/d

since they both coincides;

tanθ₁ = tanθ₂

m₁λ₁/d = m₂λ₂/d

multiply both sides by d

so,

m₁/m₂ = λ₂/λ₁

we substitute

m₁/m₂ = 510 nm / 425 nm

m₁/m₂ = 510 nm / 425 nm

divide through by 85

m₁/m₂ = 6 / 5

hence m₁ and m₂ are 6 and 5

so, from the previous formula

tanθ₂ = m₂λ₂/d

we substitute

tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m

tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m

tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m

tanθ₂ = 0.00145714

θ₂ = tan⁻¹( 0.00145714 )

θ₂ = 0.0834879⁰

Therefore, the required angle is 0.0834879⁰

Answer:

D) Q = 80 C

Explanation:

Given;

current flowing in the light bulb, I = 2A

time of current flow, t = 40,000 ms = 40,000 x 10⁻³ s = 40 s

The quantity of the charge is calculated as;

Q = It

where;

Q is the quantity of the charge (Coulombs)

Q = (2 ) x (40)

Q = 80 C

Therefore, the quantity of charge flowing in the circuit is 80 C

D) Q = 80 C

Answer:

Therefore, the moment of inertia is:

[tex]I=0.37 \: kgm^{2} [/tex]

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex] (1)

Where:

Let's solve the equation (1) for I

[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex]

[tex]I=Mgd(\frac{T}{2\pi})^{2}[/tex]

Before find I, we need to remember that

[tex]T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s[/tex]

Now, the moment of inertia will be:

[tex]I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}[/tex]  

Therefore, the moment of inertia is:

[tex]I=0.37 \: kgm^{2} [/tex]

I hope it helps you!

Answer:

Energy = .13 W / m^2      energy of incident energy

N = 3500 Watts / day       power needed

N = 3500 Watts  (3600 * 24 sec) = .0405 Watts/sec

The problem must mean that one needs 3.5 Kw-days

3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules

150 J/sec-m^2 * .13 = 19.5 J / sec-m^2      usable energy

In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2    usable energy received

Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2

One would need 180 m^2 of solar panels

That's quite a lot of energy

A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.

Answer:

b. The heavier one reaches the bottom first.

Answer:

B

Explanation:

The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]

Calculating the common potential:

[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]

Calculating the charge after redistribution:

[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]        

[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]

Answer:

[tex]h=10m[/tex]

Explanation:

From the question we are told that:

Area [tex]a=70 x 10^{-6}[/tex]

Force [tex]F=7N[/tex]

Generally the equation for Pressure is mathematically given by

Pressure = Force/Area

[tex]P=\frac{F}{A}[/tex]

[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]

[tex]P= 1*10^{5} Pa[/tex]

Generally the equation for Pressure is also mathematically given by

[tex]P=hpg[/tex]

Therefore

[tex]h=\frac{P}{hg}[/tex]

[tex]h=\frac{10000}{1000*9.8}[/tex]

[tex]h=10m[/tex]

Answer:

to the right

Explanation:

if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.

Answer:

s=50m

Explanation:

you can use the formula

s=ut+1/2at²

s=0t+1/2(1)10²

=1/2(100)

=50

I hope this helps

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because any object at rest or in uniform motion has no speed or velocity

Answer:

0.032 [tex]m/s^2[/tex]

Explanation:

Given :

Weight of the astronaut = 69 kg

Weight of the physics book = 28 kg

Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]

The force that is applied on the astronaut :

[tex]F=ma[/tex]

   [tex]$=69 \times 0.013$[/tex]

   = 0.897 N

Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N

Therefore, the acceleration of the physics book is given by :

[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]

[tex]$a = \frac{0.897}{28}$[/tex]

a = 0.032 [tex]m/s^2[/tex]

Hence, the acceleration of the physics book is  0.032 [tex]m/s^2[/tex].

Answer:

The acceleration of astronaut is 5.27 x 10^-3 m/s^2.

Explanation:

mass of astronaut, M = 69 kg

Mass of book, m = 28 kg

acceleration of book, a = 0.013 m/s^2

Let the acceleration of astronaut is A.

According to the Newton's third law, for every action there is an equal and opposite reaction.

So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.

M A = m a

69 x A = 28 x 0.013

A = 5.27 x 10^-3 m/s^2

Answer:

The distance is 22.45 cm.

Explanation:

Height of cylinder, H = 14 cm

depth of hole, h = 5 cm

The distance of landing of stream from the base of cylinder is

[tex]R = 2\sqrt{H(H-h)}\\\\R = 2\sqrt{14(14-5)}\\\\R = 22.45 cm[/tex]

Answer:

6.08

Explanation:

Given that,

Current, I = 0.8 A

Time, t = 76 h

We need to find the Ah rating of a battery. It can be calculated by taking the product of current and time. So,

Ah = (0.8)(76)

= 6.08 Ah

So, the Ah rating of the battery is 6.08.

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

[tex]U = 0.5 C V^2[/tex]

The capacitance is given by

[tex]C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J[/tex]

Answer:

The resolving power remains same.

Explanation:

The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.

As the diameter is same but the focal length is doubled so the resolving power remains same.