A wire, 0.60 m in length, is carrying a current of 2.0 A and is placed at a certain angle with respect to the magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field

Answers

Answer 1

Answer:

[tex]\theta=30 \textdegree[/tex]

Explanation:

From the question we are told that:

Current [tex]I=2.0A[/tex]

Length [tex]L=0.60m[/tex]

Magnetic field [tex]B=0.30T[/tex]

Force [tex]F=0.18N[/tex]

Generally the equation for Force is mathematically given by

[tex]F = BIL sin\theta[/tex]

[tex]sin\theta=\frac{F}{BIL}[/tex]

[tex]\theta=sin^{-1}\frac{0.18}{0.3*2*0.6}[/tex]

[tex]\theta=30 \textdegree[/tex]


Related Questions

Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?

Answers

Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

g A small object of mass 2.5 g and charge 18 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field

Answers

Answer: [tex]1361.11\ N/C,\text{upward}[/tex]

Explanation:

Given

Mass of particle is [tex]m=2.5\ gm[/tex]

Charge of particle is [tex]q=18\ \mu C[/tex]

Electrostatic force must balance the weight of the particle

[tex]\lim_{n \to \infty} a_n \Rightarrow mg=qE\\\\\Rightarrow E=\dfrac{2.5\times 9.8\times 10^{-3}}{18\times 10^{-6}}\\\\\Rightarrow E=1361.1\ N/C[/tex]

Direction of the electric field is in upward direction such that it opposes the gravity force.

1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N​

Answers

Answer:

D) 1003 N​

Explanation:

Given the following data;

Mass of man = 85 kg

Acceleration of elevator = 2 m/s²

Acceleration due to gravity, g = 9.8 m/s²

To find the force exerted by the man on the floor;

Force = mg + ma

If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?

a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.

Answers

Answer: They are equal in magnitude.

- The signs of the two changes in potential energy are opposite

Explanation:

When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.

Also, it should be noted that the signs of the two changes in potential energy are opposite.

In December of 2011 they announced that a planet has been discovered in a habitable zone around a
star! It has clouds! It has twice the radius of the earth, but with the same density as earth, about 5.515 × 10^3kg/m3
. Find the new acceleration of gravity on the surface of this planet.

Answers

Explanation:

The density of earth [tex]\rho_E[/tex] is given by

[tex]\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}[/tex]

and in terms of this density, we can write the acceleration due to gravity on earth as

[tex]g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E[/tex]

Similarly, the acceleration due to gravity [tex]g_P[/tex] on this new planet is given by

[tex]g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}[/tex]

[tex]\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P[/tex]

We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite [tex]g_P[/tex] as

[tex]g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)[/tex]

[tex]\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E[/tex]

[tex]\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2[/tex]

Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.

Answers

Answer:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

Explanation:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.

If the hypothesis and the model used is correct, the relationship to be tested is

              T² =(4π² /g)   L  

by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.

You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)

Answers

Answer:

The current is more in the parallel combination than in the series combination.

Explanation:

two resistances, R1 and R2 are connected to a battery of voltage V.

When they are in series,

R = R1 + R2

In series combination, the current is same in both the resistors, and it is given by Ohm's law.

V = I (R1 + R2)

[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)

When they are connected in parallel.

the voltage is same in each resistor.

The effective resistance is R.

[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]

So, the current is

[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)

So, the current is more is the parallel combination.

The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. Express your answer numerically in joules.

Answers

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

[tex]A = 10 \ cm^2[/tex]

   [tex]$=0.0010 \ m^2$[/tex]

d = 10 mm

  = 0.010 m

Then, Capacitance,

[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]

[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]

[tex]$C=2.655 \times 10^{12} \ F$[/tex]

[tex]$U_1 = \frac{1}{2}CV^2$[/tex]

[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]

Now,

[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]

And

[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]

In parallel combination,

[tex]$C_{eq}= C_k + C_{air}$[/tex]

[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]

[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]

[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]

Then energy,

[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]

[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]

b). Now the charge on the [tex]\text{capacitor}[/tex] is :

[tex]$Q=C_{eq} V$[/tex]

[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]

[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]

Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :

[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]

[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]

[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]

[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]

Without the dielectric,

[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]

[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]

[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]

A motorcyclist start from rest to reaches 6m/s with uniform acceleration for 3s what his acceleration?​

Answers

Answer:

[tex]\boxed {\boxed {\sf 2 \ m/s^2}}[/tex]

Explanation:

Acceleration is the rate of change in velocity with respect to time. It is calculated by dividing the change in velocity by the change in time. The formula is:

[tex]a= \frac{ \Delta v}{\Delta t}[/tex]         or          [tex]a= \frac{v_f-v_i}{\Delta t}[/tex]

The change in velocity is the difference between the initial velocity and the final velocity. The motorcycle starts at rest, or 0 meters per second and reaches 6 meters per second. The change in time is 3 seconds.

[tex]\bullet \ v_f= 6 \ m/s\\\bullet \ v_i= 0 \ m/s \\\bullet \ \Delta t = 3 \ s[/tex]

Substitute the values into the formula

[tex]a= \frac { 6 \ m/s - 0 m/s}{3 \ s}[/tex]

Solve the numerator.

[tex]a= \frac{6 \ m/s}{3 \ s}[/tex]

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

The motorcyclist's acceleration is 2 meters per second squared.

b. A bird in air looks a fish vertically below it inside the water from a distance 5m from surface of water and fish lies at depth 4m from the surface of water. IF Mw= 4/3, what is the distance of fish as observed by bird?

Answers

Answer:

the distance of the fish (as  seen by the bird) is greater than the actual distance.

Reason-

it is due to the apparent depth and differences between the refractive indices.

Have a nice day!

In the lab room, you are sitting in an office chair with wheels while holding onto a force sensor, and the chair is at rest. One end of a lightweight string is attached to the force sensor, and your lab partner is holding the other end of the string. Your lab partner then moves away from you, pulling on the string. Describe how your lab partner must move for the force sensor to read a constant force. Explain

Answers

Answer:

a circular motion a constant force can be measured

Explanation:

The force is expressed by the relation

         F = m a

The bold are vectros.

Therefore, when your partner moves away, he has a reading of a force, so that this force remains constant there must be an acceleration at all times, one way to achieve these is with a circular motion with constant speed, in this case the module of the velocity is constant, but the direction changes at each point and there is an acceleration at each point.

Consequently with a circular motion a constant force can be measured

The force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.

What is Force?

A force can be defined as an influence that can change the motion of an object. The force is expressed by the relation

[tex]F = m a[/tex]

The force is dependent on the mass and acceleration of the object. The acceleration is a vector quantity, so the force will be a vector quantity.

Given that, in a lab room, you are sitting on a wheelchair at one end and at the other end, lab partner then moves away from you, pulling on the string that is attached to the force sensor.

When the lab partner moves away and pulled the string, there must be an acceleration during the motion. If the lab partner moves in a circular motion, then the velocity is constant but the direction changes at each point. There is an acceleration at each point that will be constant.  

As the force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.

To know more about the force, follow the link given below.

https://brainly.com/question/26115859.

two electrons are separated by 1.10m, What is the magnitude of the electric force each electron exerts on the other?

Answers

Answer:

4.56×10¯⁷¹ N

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 1.10 m

Force (F) =?

NOTE:

Gravitational constant (G) = 6.67×10¯¹¹ Nm² /Kg²

Mass of electron = 9.1×10¯³¹ Kg

Mass of the two elections = M₁ = M₂ = 9.1×10¯³¹ Kg

Thus, we can obtain the force of attraction between the two elections as illustrated below:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × (9.1×10¯³¹)² / (1.1)²

F = 4.56×10¯⁷¹ N

Thus, the force of attraction between the two elections is 4.56×10¯⁷¹ N

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magnetic B=B k and electric E=−E k fields with the initial velocity v=v0 i. The magnitudes of the fields: B=0.18 T, E=278 V/m, and the initial speed v0=2.1 m/s are given. Find at what time t, the particle's speed would become equal to v(t)=3.78·v0:

Answers

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

[tex]F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j[/tex]

F = ma

[tex]a ^{\to}= \dfrac{F^{\to}}{m}[/tex]

[tex]a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)[/tex]

[tex]a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j[/tex]

At time t(sec; the partiCle velocity becomes [tex]v(t) = 3.78 v_o[/tex]

The velocity of the charge after the time t(sec) is expressed by using the formula:

[tex]v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}[/tex]

3) A lead bullet initially at 30 C just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact. (Specific heat of lead is 0.128 kJ/kg K and lead latent heat of fusion is 24.7 kJ/Kg and melting point of lead is 600 K).

Answers

Answer:

The speed of bullet is 354.2 m/s

Explanation:

initial temperature, T = 30 degree C

specific heat, c = 128 J/kg K

Latent heat, L = 24.7 x 1000 J/kg

melting point = 600 K = 327 degree C

Let the mass is m and the speed is v.

Kinetic energy = heat used to increase the temperature + Heat used to melt

[tex]\frac{1}{2} mv^2 = m c (T' - T) + m L\\\\0.5 v^2 = 128 \times (327 - 30) + 24.7\times 1000\\\\0.5 v^2 = 38016 + 24700 \\\\0.5 v^2 = 62716\\\\v = 354.2 m/s[/tex]

An airplane which intends to fly due south at 250 km/hr experiences a wind blowing westward at 40 km/hr. What is the actual speed of the airplane relative to the ground? ​

Answers

Answer:

simple is rumple a daily ok I'll be

Difference between uniform motion and non uniform motion

Answers

Answer:

When an object covers equal distance in an equal interval of time, it is uniform motion but when an object covers unequal distance in an equal interval of time, it is called non uniform motion.

Two charged objects attract each other with a force 1.0 N. What happens to the force between them if one charge is increased by a factor of 2, the other charge is increased by a factor of 4, and the separation distance between their centers is reduced to 1/4 its original value

Answers

Answer:

F' = 128 N

Explanation:

The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:

[tex]F = \frac{kq_1q_2}{r^2}\\\\[/tex]

where,

F = Force of attraction = 1 N

G = universal gravitational constant

q₁ = magnitude of the first charge

q₂ = magnitude of the second charge

r = distance between charges

Therefore,

[tex]1\ N = \frac{kq_1q_2}{r^2}[/tex] --------------------- eq(1)

Now, we apply the changes given in the question:

[tex]F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})[/tex]

using eq (1):

F' = 128(1 N)

F' = 128 N

CHEGG Over the course of a multi-stage 4820-km bicycle race, the front wheel of an athlete's bicycle makes 2.40x106 revolutions. How many revolutions would the wheel have made during the race if its radius had been 1.4 cm larger?

Answers

Answer:

θ' =  14.44 × [tex]10^{6}[/tex]

Explanation:

given data

total distance is d = 4820

radius = 1.4 cm

solution

we get here total angle by which the wheel rotates traveling is express as

⇒  [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex]     ................1

and

total angle (θ)  and the total distance (d) express as

⇒ d = r × θ       ...............2

here r is radius

and here rotated through some other angle θ' so put value in given equation and find revolutions    

⇒  d = (r+r)θ'      ........3

here  r = d/θ

so

⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]    

so put value and get θ'

⇒  θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]  

⇒  θ' =  14.44 × [tex]10^{6}[/tex] rev

With respect to a right handed Cartesian coordinate system and given that . A = 4i + k and B = 2i + j _ 3k find A cross B

Answers

Using the left-hand rule,

[tex](4\,\vec\imath+\vec k)\times(2\,\vec\imath+\vec\jmath-3\,\vec k) = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\4&0&1\\2&1&-3\end{vmatrix} = -\vec\imath+14\,\vec\jmath+4\,\vec k[/tex]

Then in the right-handed rectangular coordinates, the cross product is the negative of this,

[tex]\boxed{\vec\imath-14\,\vec\jmath-4\,\vec k}[/tex]

A cylindrical 5.00-kg reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel. The bucket starts from rest and falls for 4.00 s.

Required:
a. What is the linear acceleration of the falling bucket?
b. How far does it drop?
c. What is the angular acceleration of the reel?

Answers

I think the answer is C

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to the potential difference between the plates as they are being separated?
A) It increases.
B) It decreases.
C) It remains constant.
D) It cannot be determined from the information given.

Answers

The correct answer is A) it increases

What is the incorrect statement regarding the isotopes of the same element?
1) Electronic configuration is equal
2) Mass number is equal
3) Number of protons are equal
4) Number of electrons are equal​

Answers

Answer:

1231

Explanation:

What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.

1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J

Answers

Frequency =v=3.6×10^15Hz

We know

[tex]\boxed{\sf E=hv}[/tex]

[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]

[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]

[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]

[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]

Answer:

D!!!!!

Explanation:

a cell phone is released from the top with the speed of 10ms what is the speed 3s after?

Answers

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 600 lines per mm, projecting a pattern on a screen 2.9 m behind the grating.

Required:
How wide is the spectrum corresponding to m=1?

Answers

Answer:

  Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

        a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

   a = 2 lines 1/600

   a = 2/600

    a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

      tan θ = y / L

as the measured angles are small

      tan θ = sin θ / cos θ sin θ

      sin θ = y / L

we substitute

     a  y/L = λ

     y = λ L / a

for λ = 400 10-9 m

      I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

      i = 346.89 10⁻⁶ m

f

or λ = 700 nm

        y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

        y_f = 609.609 10⁻⁶ m

the separation of this spectrum

        Δr = v_f - i

        Dr = (609.609 - 346)  10 ⁻⁶

        Dr = 263 10⁻⁶ m

Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.

Answers

Answer:

5.71×10¹⁴ Hz

Explanation:

Applying,

v = λf................. Equation 1

Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency

make f the subject of the equation

f = v/λ............. Equation 2

From the question,

Given: λ = 525 nm = 5.25×10⁻⁷ m,

Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s

Substitute these values into equation 2

f = (3.0×10⁸)/(5.25×10⁻⁷)

f = 5.71×10¹⁴ Hz

Hence the frequency of light is 5.71×10¹⁴ Hz

Identify the factors that affect the intensity of radiation detected from a radioactive source. Select one or more: The color of the source Type of emission from the source Distance of the detector from the source Type of materials between the source and the detector

Answers

The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.

The factors that affect the intensity of radiation are

Type of emission from the source :This  can be alpha, gamma, beta or electromagnetic rays etc

Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.

Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.

Read more on Radiation Intensity here:  https://brainly.com/question/10148635

A capacitor is connected to an ac generator that has a frequency of 3.2 kHz and produces a rms voltage of 2.0 V. The rms current in the capacitor is 28 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.7 kHz, the rms current in the capacitor is 70 mA. What rms voltage does the second generator produce

Answers

Answer:

The rms voltage of new generator is 3.4 V.

Explanation:

f = 3200 Hz

rms voltage, V = 2 V

rms current, i = 28 mA

Now

f' = 4700 Hz

rms current, i' = 70 mA

let the new rms voltage is V'.

[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]

From the given picture What's the force? And where did it happen? (at least 2 forces)

Answers

Answer:

the force happens on the wall and couch

Explanation:

she is using her arm strength to lift and hold

Describe how the words Science and optics would appear when viewed in a plane mirror?

Answers

Answer:

Lateral inversion will occur in a plane mirror.

Explanation:

When words are displayed in a plane or flat mirror, the result is that if the words are displayed left, they change to right and if they were normally displayed right, they change to left. This phenomenon is known as lateral inversion. So, this will apply to the words, Science and optics. Only the sides will be interchanged.

A plane mirror reflects light, therefore, the image that is produced by it remains the same size. The image produced will not appear upside down. Only the sides will be interchanged.

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