A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
(a) 0.30 s V
(b) 3.0 s V
(c) 65 s V

Answer:

resultant vector =0

Explanation:

because it is connected head to tail in a closed figure

Answer:

[tex]K=0.023J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.15[/tex]

Velocity [tex]v=0.5m/s[/tex]

Angular Velocity [tex]\omega=8.4rad/s[/tex]

Generally the equation for Kinetic Energy is mathematically given by

[tex]K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)[/tex]

[tex]K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))[/tex]

[tex]K=0.023J[/tex]

Answer:

the water jet and the negative object attract

Explanation:

Let's analyze the situation. Water is a good conductor of electricity, so when an object with a positive charge is brought closer, a charge of the opposite sign is created, which is why the two objects attract each other. The charge created comes from the ground

When claiming the object, the charge is reduced to zero or the charge goes to earth since water is a good conductor of electricity, therefore when approaching an object with a negative charge, more charges go to earth and the ring is left with a positive charge and attracts it to the object.

In short, the water jet and the negative object attract

Answer:

C Quincy

Explanation:

Both Frieda and Vladimir are too young to be making their own decisions. Lucinda has limited freedom due to having two children, while Quincy is just now becoming an adult and has his whole life still ahead of him. Therefore, Quincy has the most freedom to make their own lifestyle decisions.

I hope this helps!

Answer:

The correct answer is C. Quincy is 16 years old and lives in a dormitory

Explanation:

Quincy being 16 may be able to get a job and hold a membership at a gym even if not at a gym he still has the most freedom period with Lucinda having children and Frieda and Vladimir being to young to make the choices of exercising on their own.

Please tell me if I'm wrong so I may give you the correct answer!

Happy Holidays!!  

Answer:

e

Explanation:

From the given information:

Suppose Q = total charge of the sphere.

here, the electric field outside the sphere at distance R/2 can be expressed as:

[tex]E_1 = \dfrac{1}{4 \pi \varepsilon _o}* \dfrac{Q}{(R + \dfrac{R}{2})^2}[/tex]

where:

[tex]k = \dfrac{1}{4 \pi \varepsilon _o}[/tex]

[tex]E_1 = \dfrac{kQ}{(\dfrac{3R}{2})^2}[/tex]

[tex]E_1 = \dfrac{4kQ}{9R^2}[/tex]

For the electric field inside the sphere, we have:

[tex]E_2 = \dfrac{kQr}{R^3}[/tex]

here:

r = distance of the point from the center = R/2

R = radius of the sphere

∴

[tex]E_2 = \dfrac{kQ * \dfrac{R}{2}}{R^3}[/tex]

[tex]E_2 = \dfrac{kQ }{2R^2}[/tex]

As such, the ratio of the electric field outside the sphere to the one inside is:

[tex]\dfrac{E_1}{E_2} = \dfrac{ \dfrac{4kQ}{9R^2}}{ \dfrac{kQ }{2R^2}}[/tex]

[tex]\dfrac{E_1}{E_2} = \dfrac{4kQ}{9R^2} \times \dfrac{ 2R^2 }{kQ}[/tex]

[tex]\mathbf{\dfrac{E_1}{E_2} = \dfrac{8}{9}}[/tex]

For a solid uniformly charged sphere of radius R, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is -  e) 8/9

field  [tex]E2= \frac{k(Q)}{(3R/2)^2}[/tex]

=  [tex]\frac{4}{9} \frac{k(Q)}{R^2}[/tex]

=

Thus, E2/E1

= (4/9)/(1/2)

= 8/9

Thus, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is -  8/9

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Answer:

microsecond = 1 × 10-6 seconds

1 second = 1 × 100 seconds

1 microsecond = (1 / 1) × 10-6 × 10-0 seconds

1 microsecond = (1) × 10-6-0 seconds

1 microsecond = (1) × 10-6 seconds

1 microsecond = 1 × 1.0E-6 seconds

1 microsecond = 1.0E-6 seconds,just do like it

Answer:

6.3 x [tex]10^{-3}[/tex]

Explanation:

1 millisecond = 1000 microseconds

0.0063 millisecond = 6.3 microseconds

Answer:

Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.

If the cannon throws the banana with the same force the monkey falls

(m.g=Fz <=> m.9,81N/kg=...N).

Then the throw will slow down because of the gravitational pull.

Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.

The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.

If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.

So to do so you need to throw the bananas with a speed of at least 9,81m.s

Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.

I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.

kind regards

Answer:

Explanation:

so opposite and equal , right?    forces are. soooo..

528+52= 580 N    is the force that is being exerted up on the scale

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

Answer:

The spring constant of the spring is 10.3 N/m.

Explanation:

Given that,

Mass of a bungee jumper, m = 59 kg

The period of oscillation, T = 0.25 min = 15 sec

We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :

[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]

Where

k is the spring constant

[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]

So, the spring constant of the spring is 10.3 N/m.

Answer:

a. Total displacement = 140 km/h

b. Average velocity = 35 km/h

Explanation:

Given the following data;

Average velocity A = 80 km/h

Time A = 2.5 hours

Average velocity B = 40 km/h

Time B = 1.5 hours

a. To find the total displacement for the 4 h trip;

Total time = Time A + Time B

Total time = 2.5 + 1.5

Total time = 4 hours

Next, we would determine the displacement at each velocity.

Mathematically, displacement is given by the formula;

Displacement = velocity * time

Substituting into the formula, we have;

Displacement A = 80 * 2.5

Displacement A = 200 km/h

Displacement B = 40 * 1.5

Displacement B = 60 km/h

Total displacement = Displacement A - Displacement B

Total displacement = 200 - 60

Total displacement = 140 km/h

b. To find the average velocity for the total trip;

Mathematically, the average velocity of an object is given by the formula;

[tex] Average \; velocity = \frac {total \; displacement}{total \; time} [/tex]

Substituting into the formula, we have;

[tex] Average \; velocity = \frac {140}{4} [/tex]

Average velocity = 35 km/h

Answer:

1. The electric potential would stay the same.

2. The electric potential energy would double

Explanation:

1. Since the electric potential at the point is V and is not dependent on the charge q, the electric potential at the point still remains the same even if the charge q is doubled to 2q.

So, the electric potential remains the same when the charge is doubled to 2q.

2. Since potential energy U = qV where q = charge and V = electric potential. So when the charge is q and electric potential, V, the electric potential energy at the point is U = qV.

When the charge is doubled to 2q and the electic potential remains the same, the electric potential energy is now U' = 2qV = 2U (since U = qV).

So, the electric potential energy at the point is doubled when the charge is doubled.

The period and the amplitude of the weight suspended from spring are 0.33 seconds and 10 centimeters, respectively.

1) The period is given by:

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

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

[tex] T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s [/tex]

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

[tex] A = \frac{20 cm}{2} = 10 cm [/tex]  

Therefore, the amplitude is 10 cm.          

You can learn more about the period and amplitude here: https://brainly.com/question/15169209?referrer=searchResults

I hope it helps you!                    

Solution-1:-

[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]

Solution:-2

[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]

Solution:-3

[tex]\boxed{\sf 320m}[/tex]

Solution:-4

[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]

Solution:-5

[tex]\boxed{\sf Mg_3N_2}[/tex]

Solution:-6

[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]

Solution:-7

[tex]\boxed{\sf {}^{}_{}N}[/tex]

Solution:-8

[tex]\boxed{\sf Galactuse}[/tex]

Solution:-9

[tex]\boxed{\sf Y-X}[/tex]

Solution:-10

[tex]\boxed{\sf Cell\:wall}[/tex]

Answer:

Option (c).

Explanation:

Let the mass of each cart is m and the force is F.

Time for cart A is 2t and for cart B is t.

Work done is given by the

W= force x displacement

As the distance is given by

S= u t +0.5 at^2

So, when the time is doubled the distance is four times.

So, WA = F x 4 S

WB = F x S

WA= 4 WB

Answer:

imma try and still get wrong

The fielder's required speed is (D + (v * t * cos(a))) / t.

To catch the ball, the fielder must run at a speed that allows them to cover the horizontal distance D between themselves and the hitter, as well as the distance the ball travels horizontally before hitting the ground. The time it takes for the ball to hit the ground can be determined based on its initial vertical velocity and the acceleration due to gravity.

Let's call the time the ball is in the air "t." The horizontal distance the ball travels is given by D + (v * t * cos(a)), where "v" is the initial speed of the ball and "a" is the angle of the ball's flight above ground. The fielder's running speed should equal this horizontal distance divided by "t" to ensure they reach the ball's landing point just in time to catch it. Thus, the fielder's required speed is (D + (v * t * cos(a))) / t.

To know more about horizontal distance, here

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#SPJ2

The vertical acceleration of the projectile is at 0 m/s while the horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration  ( i.e. statement in the question is False )

Projectile motion follows a parabolic path with x and y components of its velocity and acceleration. also the acceleration of a projectile is subject only to the acceleration due to gravity unlike other kinds of motions.

In a parabolic motion an object ( projectile ) is thrown into the air and left to move through a parabolic path under the effect of acceleration due to gravity.

Hence we can conclude that the statement is false, because horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration

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[tex]\implies F_1 < F_2[/tex]

[tex] \implies F_{net} = F_2 - F1[/tex]

[tex]\implies F_{net} = 42 -15[/tex]

[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]

The net force on the 4.0 kg box is 27 N towards EAST.

Answer:

The scientific notation of 738 is 7.38 x 1002.

Answer: [tex]P=5573.43\ W[/tex]

Explanation:

Given

Mass of the elevator is [tex]M=650\ kg\\\[/tex]

Time period of ascension [tex]t=3\ s[/tex]

cruising speed [tex]v=1.75\ m/s[/tex]

Distance moved by elevator during this time

Suppose Elevator starts from rest

[tex]\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s[/tex]

Distance moved

[tex]\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5\times 0.5833\times (3)^2\\\Rightarrow h=2.62\ m[/tex]

Gain in Potential Energy is

[tex]\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N[/tex]

Average power during this period is

[tex]\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W[/tex]

Answer:

The power is 331.7 W.

Explanation:

mass, m = 650 kg

time, t= 3 s

initial  velocity,  u = 0 m/s

final velocity, v = 1.75 m/s

(a) The power is defined as the rate of doing work.

Work is given by the change in kinetic energy.

W = 0.5 m (V^2 - u^2)

W = 0.5 x 650 x 1.75 x 1.75 =  995.3 J

The power is given by

P = W/t = 995.3/3 = 331.7 W

Answer:

Approximately 2*10^13 electrons must be transferred

Explanation:

Below is the given information:

First object carries charge = -8.5 µC

Number of electrons in 1st = 8.5 x 10^-6/(1.6 x 10^-19) = 5.3125 x 10^13

Second object carries a charge = -2.0 µC

The number of electrons in 2nd = 2*10^-6/(1.6*10^-19) = 1.25 x 10^13

so, approximately 2 x 10^13 electrons must be transferred

Answer:

una ess abola cola sola answer

Answer:

No, just because the electric field is zero at a particular point, it does not necessarily mean that the electric potential is zero at that point. ... At the midpoint between the charges, the electric field due to the charges is zero, but the electric potential due to the charges at that same point is non-zero.

Explanation:

Answer:

a)  T = -22796.5 N,  b)  F = 3000 N

Explanation:

a) For this part we use Newton's second law

Let's set a reference frame with the x-axis in the direction of motion and the y-axis in the vertical direction.

We make a free-body diagram for each body,

the tractor has the bar force (T) and the push force (F) and the friction force (fr₁)

Y axis

        N₁ -W₁ = 0

        N₁ = M₁ g

X axis

        F - T - fr₁ = M₁ a

the Trailer has the bar force (T) and the friction force (fr₂)

Y axis  

      N₂ - W₂ = 0

      N₂ = m₂ g

X axis

     T - fr₂ = m₂ a

let's write the system of equations

      F - T - fr₁ = M₁ a          (1)

           T - fr₂ = m₂ a

we solve

       F - (fr₁ + fr₂) = (M₁ + m₂) a

indicate that the total friction forces are fr = 3000N

       fr = fr₁ + fr₂

        F =[tex]\frac{(M_1+m_2) a}{fr}[/tex]

let's calculate

        F =[tex]\frac{(2000+1500) \ 3}{3000}[/tex]

        F = 3.5 N

The friction force is

       fr = μ N

the norm of the system is N = N₁ + N₂

       μ = [tex]\frac{fr}{N_1 + N_2}[/tex]

       μ = [tex]\frac{3000}{2000+1500}[/tex]

       μ = 0.858

with this value we can find the friction force 1 and substitute in equation 1

       F - T - μ N₁ = M₁ a

       T = F - M₁ (a + μ g)

        T = 3.5 - 2000 (3 + 0.858  9.8)

        T = -22796.5 N

b) when the system moves with constant velocity the acceleration is zero

            F - T - fr₁ = 0

                  T - fr₂ = 0

we solve

           F + (fr₁ + fr₂) = 0

           F = fr₁ + fr₂

           F = 3000 N

Answer:

(a) 31.44 m/s (b) 164.74 m/s²

Explanation:

Given that,

The diameter of a disk, d = 12 cm

Radius, r = 6 cm

Angular speed = 5.24 rad/s

(a) Linear speed,

[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]

(b) Centripetal acceleration,

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]

Answer: [tex]3.32\times 10^4\ kN/C[/tex]

Explanation:

Given

Charge at the center of the sphere is [tex]q=2\ nC[/tex]

Charge distributed over the entire sphere [tex]Q=1\ nC[/tex]

Radius of sphere [tex]r=4.8\ cm[/tex]

Using Guass law

[tex]\Rightarrow \oint \vec{E}\cdot \vec{dA}=\dfrac{q_{enc}}{\epsilon_o}\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+\dfrac{Q}{4\pi R^2}\times 4\pi r^2)\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+Q\dfrac{r^2}{R^2})\\\\\Rightarrow E\cdot (2.4\times 10^{-2})^2=9\times 10^9(2+1\cdot \dfrac{1}{4})\times 10^{-9}\\\\\Rightarrow E=3.32\times 10^4\ kN/C[/tex]

Answer:

     a = 6.1 m / s²

Explanation:

For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle

Let's find the relative initial velocity of the two vehicles

          v₀ = v₀₂ - v₀₁

          v₀ = 25.4 - 13.6

          v₀ = 11.8 m / s

the fastest vehicle

          x = v₀ t + ½ a t²

The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is

          x = 11.4 m

let's use the expression

           v² = v₀² - 2 a x

how the vehicle stops v = 0

          a = v₀² / 2x

          a = [tex]\frac{11.8^2}{2 \ 11.4}[/tex]

          a = 6.1 m / s²

this velocity is directed to the left

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN