a simle arrangement by means of which emf are compared is known as

The double-slit experiment, fringes are more widely spaced when illumination is with monochromatic is low frequency light

What is low frequency light?

Light waves have a frequency that is proportional to their energy: Low frequency light is less energetic than high frequency light. Gamma rays therefore have the highest energy, which contributes to their high hazard to humans, whereas radio waves have the lowest energy.

Since the letters in the word "VIBGYOR" are written in decreasing frequency sequence, we can see that Orange has the lowest frequency among the alternatives while Red generally has the lowest frequency.

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Answer:

It is (B) Stimuli must be sensed in order to produce a response.

Explanation:

Answer:

A. The rays can change molecules and atoms in the body into ions.

Answer:

Endothermic

Explanation:

The arrow be released at an angle of 18 degrees to hit the bull's-eye if its initial speed is 36.0 m/s. The arrow go over the branch.

The path of the arrow is projectile when it is launched by the archer.

The horizontal distance to be covered by the arrow is 77 meters.

The initial velocity is 36m/s while releasing.

(a) We can use the formula,

R = U²sin2A/g

Where,

U is the initial velocity of leopard,

A is the angles at which the leopard jumped.

g is acceleration due to gravity having value 9.8m/s².

Putting all the values,

77 = (36)²sin(2A)/9.8

Sin(2A) = 77×9.8/36×36

Sin(2A) = 0.58

2A = 36°

A = 18°

The angles at which the arrow should be launcher is 18°.

(b) The length of the branch of the tree is 3.5 meters.

Whether is should come it the way or not,

We have to find the height of the arrow.

H = U²Sin(A)/2g.

Where H is the height,

H = 36×36Sin(18°)/2×9.8

H = 36×36×0.3/2×9.8

H = 19.38 meters.

The height of arrow is more than of the height of branch of tree. So it will go over the branch.

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The change in momentum of Kara's car is -1.43 × 10⁴ kg.m/s, the magnitude of force is -1.021 × 10⁵ N.

Force is the external agent which causes the motion of an object or it is the resistant which makes the object come at rest from motion. It is a vector quantity, because it has both the magnitude and direction.

Mass of Kara's car = 1300 Kg

moving with speed = 11 m/s

time taken to stop = 0.14 s

final velocity = 0 m/s

distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

Δ P = m Δ v          

Δ P = m(vf - vi)

Δ P = 1300 (0 - 11)

Δ P = -1.43 × 10⁴ kg.m/s

The impulse is equal to the change in momentum of the car

I = -1.43 × 10⁴ kg.m/s

Magnitude of the force experienced by Kara

I = F × t

where, I is impulse acting on the car

t is time

- 1.43 × 10⁴ = F × 0.14

F = -1.021 × 10⁵ N

Negative sign represents the direction of the force.

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Your question is incomplete, most probably the complete question is:

Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was stopped in her lane 30 feet ahead, Kara rear-ended Lisa's rented Taurus. Kara's 1300-kg car was moving at 11 m/s and stopped in 0.14 seconds.

a. Determine the momentum change of Kara's car.

b. Determine the impulse experienced by Kara's car.

c. Determine the magnitude of the force experienced by Kara's car.

The correct equation for the x-axis of object A is F-T = mga, where F is the force acting on object A, T is the tension in the rope, m is the mass of object A, g is the acceleration due to gravity, and a is the acceleration of object A.

From NA - WA = 0, we know that the normal force acting on object A is equal to its weight, NA = WA = mg = 75 kg * 9.81 m/\(s^2\) = 735.75 N.

From Ng - WB = 0, we know that the weight of object B is equal to the weight of the hanging mass, WB = mg = 75 kg * 9.81 m/\(s^2\) = 735.75 N.

The tension in the rope is equal to the weight of the hanging mass, T = WB = 735.75 N.

From F - T = mga, we can solve for the acceleration of object A, a = (F - T) / m = (mA * g - T) / mA = (mA * g - 735.75 N) / mA.

Substituting the given values of mA = 50 kg and g = 9.81 m/\(s^2\), we get a = (50 kg * 9.81 m/\(s^2\) - 735.75 N) / 50 kg = 4.549 m/\(s^2\).

Therefore, the correct equation for the x-axis of object A is F-T = mga = 50 kg * 4.549 m\(s^2\) = 227.45 N.

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The current in the circuit when the voltage across the resistor is equal to the voltage across the inductor is given by:

i(t) = [V_B / R] (1 - e^(-Rt/L)) where t is the time after the switch is closed.

The series circuit in the figure contains an ideal battery with a constant terminal voltage V_B=52 V, an ideal inductor L=42 H, a R=20-ohm resistor, and a switch S. Initially, the switch is open, and there is no current in the inductor. At time t=0 s, the switch is suddenly closed. What is the current in the circuit when the voltage across the resistor is equal to the voltage across the inductor-If the voltage across the resistor and the voltage across the inductor are equal, then the voltage across the inductor is V = IR, where I is the current in the circuit.

According to Kirchhoff's law, the voltage across the resistor is equal to the voltage of the battery minus the voltage across the inductor, i.e., V = V_B - L (di / dt), where di/dt is the rate of change of current, and L is the inductance of the inductor. Equating these two expressions for V, we have: IR = V_B - L (di / dt). Rearranging this equation gives us the following differential equation:di / dt + (R/L) i = (V_B / L)The solution of this differential equation is given by: i(t) = [V_B / R] (1 - e^(-Rt/L)).

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(a) The  force pulling the apple is the weight of the apple and the magnitude is determined as 0.98 N.

(b) The gravity needed is 0.98 N.

(c) The energy transfer that takes place is 5.88 J.

The gravity needed is calculated as follows;

Fg = mg

where;

Fg = 0.1 x 9.8

Fg = 0.98 N

The  force pulling the apple is equal in magnitude to force of gravity and the magnitude is determined as 0.98 N.

So the pulling force on the apple is the weight of the apple.

The energy transfer that takes place is calculated as follows;

E = P.E = mgh

E = 0.98 N x 6 m

E = 5.88 J

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The weight of the Mars Rover Curiosity on Earth and on Mars is 8820 N and 3330 N respectively.

The weight of an object is given by the product of its mass and the gravitational field strength at its location.

On Earth:

On Mars:

Therefore, the weight of the Mars Rover Curiosity on Earth and on Mars are 8820 N and 3330 N respectively.

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Chemical change occurs when one or more substances undergo a chemical reaction, resulting in the formation of a new substance with different physical and chemical properties.

During a chemical change, the original substance is transformed into a different substance with different chemical composition and characteristics.

Some common signs of a chemical change include the release or absorption of heat, light, or gas; a color change; the formation of a precipitate (a solid that separates from a liquid mixture); or a change in the chemical properties of the substance, such as its acidity or reactivity.

Chemical change is different from physical change, which involves a change in the physical properties of a substance without changing its chemical composition. Examples of physical changes include changes in state (solid, liquid, gas), changes in shape or size, and changes in density or temperature.

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To find:-

Answer:-

We are here given that two long current carrying wires are having same current. We need to find out the magnetic field at the centre between the wires .

We know that for a point between two ends of a wire , magnetic field is given by,

\(\implies B =\dfrac{\mu_0}{4\pi}\dfrac{2i}{d}\\\)

where ,

Now since magnetic field is a vector quantity we need to find out the direction of the field . We can do so by using Right Hand thumb rule .

Right hand thumb rule :-

Hold the wire , in your hand with thumbs towards the direction of the current, then the curling of the fingers would give you the direction of the magnetic field.

For wire AB :-

The direction comes to be down the page .

For wire CD :-

The direction comes to be down the page .

Calculating net magnetic field:-

The net magnetic field will be the sum of both the fields .

\(\implies B_{net}=\dfrac{\mu_0}{4\pi}\dfrac{2i}{d}+\dfrac{\mu_0}{4\pi}\dfrac{2i}{d} \\\)

\(\implies B_{net}=\dfrac{\mu_0}{4\pi}\dfrac{4i}{d}\\\)

\(\implies \underline{\underline{\green{ B_{net}=\dfrac{\mu_0i}{ \pi d}}}}\\\)

The direction is down the page .

and we are done!

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

A, Passive Solar.

Explanation:

Answer:

7bright fringes

Explanation:

First we know that

Sinစ m,max= Mmax, x lambda/d

To find Mmax

Sin90° = Mmax x lambda/d = 1

So

Mmax= d/ lambda

= lambda= 500nm

d= 1/550mm

So

= 1.8 x10-6/500x 10^-9m

= 3.6

So m is approx 3

So 3 bright fringes above below and the central bright fringe itself making 7bright fringes

Answer:

360000

Explanation:

Answer:

360000 milliseconds

Explanation:

I have this remembered dont worry homie i got chu

Explanation:

Given that,

Mass of a puck, m = 8 kg

Initial velocity, u = 10i m/s

Final velocity, v = (10i+12j) m/s

Time, t = 8 s

(a) Force acting on an object is given by the product of mass and acceleration.

F = ma

a is acceleration

\(F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{8\times (10i+12j-10i)}{8}\\\\F=12j\ \text{N}\)

(b) The magnitude of force is given by :

\(F=\sqrt{12^2+0^2} \\\\F=12\ N\)

Hence, the component of force is 12j N and its magnitude is 12 N.

The magnitude of the masses in planes B and E is 40 kg, and their positions are 120° and 240°, respectively, from the reference point on the shaft to achieve complete rotating balance.

To achieve complete rotating balance, the sum of the moments of the masses in planes A, C, D, B, and E should be equal to zero. Let's determine the magnitude of the masses in planes B and E and their positions.

Consider the moments of the masses in planes A, C, and D. The moment of a mass is given by the product of its magnitude and the sine of the angle between the mass and a reference line. The moments of masses A, C, and D are:

Moment of A = 50 kg * sin(0°) = 0 kg·m,

Moment of C = 40 kg * sin(90°) = 40 kg·m,

Moment of D = 80 kg * sin(135°) = -80 kg·m.

Since the moments of A, C, and D are known, we can use the principle of complete rotating balance to determine the magnitude and position of the masses in planes B and E.

Let's assume the magnitude of the masses in planes B and E as M. The moments of masses B and E can be represented as:

Moment of B = M * sin(120°) = M * √(3)/2,

Moment of E = M * sin(240°) = -M * √(3)/2.

Using the principle of complete rotating balance, the sum of the moments should be zero. Thus, we have:

Moment of A + Moment of C + Moment of D + Moment of B + Moment of E = 0.

0 + 40 kg·m + (-80 kg·m) + M * √(3)/2 + (-M * √(3)/2) = 0.

Simplifying the equation:

40 kg·m - 80 kg·m + M * √(3)/2 - M * √(3)/2 = 0,

-40 kg·m = 0.

From the equation, we can deduce that M must be equal to 40 kg to satisfy the condition of complete rotating balance.

Finally, we determine the positions of masses B and E. Since planes A, C, D, B, and E are equidistant from one another, and the angle between A and C is 90°, we divide the circle into 360°/5 = 72° sections. Thus, the positions of masses B and E are:

Position of B = 0° + 2 * 72° = 144°,

Position of E = 0° + 4 * 72° = 288°.

Therefore, the magnitude of the masses in planes B and E is 40 kg, and their positions to put the shaft in complete rotating balance are 144° and 288°, respectively, from the reference point on the shaft.

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Answer:

Explanation:

The horizontal velocity is 1.61.

the frictional force on the broom has magnitude μ * m * g * cos(theta)

The frictional force on the broom can be determined using the equation:

F_friction = μ * F_norm

Where F_friction is the frictional force, μ is the coefficient of friction, and F_norm is the normal force. The normal force is equal to the force exerted on the broom perpendicular to the surface of the floor. Since the broom is being pushed across a horizontal surface, the normal force is equal to the force of gravity, which is equal to the mass of the broom (m) multiplied by the acceleration due to gravity (g).

F_norm = m * g

The force being applied to the broom is 7N and it is at angle theta with the horizontal. The component of this force perpendicular to the surface of the floor is the force that acts on the normal force.

F_norm = 7 * cos(theta)

Therefore,

F_friction = μ * m * g * cos(theta)

So, the frictional force on the broom has magnitude μ * m * g * cos(theta)

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Answer:

256 N

Explanation:

formula of centripetal force = mv²/r

m= 2kg

v= 8m/s

r= 0.5m

mv²/r = 2×8²/0.5 = 256N

Answer:

Explanation:

1. Place a cup of hot tea on a flat surface.

2. Place a thermometer in the tea and record the temperature.

3. Place a fan in front of the cup of tea and turn it on.

4. Place the thermometer in the tea again and record the temperature.

5. Compare the two temperatures and observe the difference.

6. The difference in temperature is an indication of the thermal energy that has been dissipated from the cup of hot tea.

The friction of a 5 kg block that is on a table when a 15 N force is exerted on the side of the block, the acceleration is 2 m/s² is 5 N

∑F = m a

F - f = m a

F = Applied force

f = Frictional force

m = Mass

a = Acceleration

F = 15 N

m = 5 kg

a = 2 m / s²

f = F - m a

f = 15 - ( 5 * 2 )

f = 15 - 10

f = 5 N

Frictional force is a type of force that resists the motion of an object. There are two types of frictional force. They are:

Therefore, the friction of the block is 5 N

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(a) The induced current in the circuit is clockwise.

b. To find the power (P) using P = Fd/t or P = Fv (since d/t = v). Here, F = ILB (from the Lorentz force), so P = (ILB)v.

(a) The induced current in the circuit is clockwise.

This can be determined using the right-hand rule.

As the metal bar moves to the left through the magnetic field directed out of the plane, the generated force on the electrons (Lorentz force) will push them toward the top rail, creating a clockwise current.

(b) To find the rate at which the applied force is doing work on the bar, first calculate the induced EMF (ε) using Faraday's law:

induced EMF (ε) using Faraday's law:

ε = BLv

= (0.65 T) * (0.36 m) * (5.9 m/s)

= 1.389 Tm²/s

= 1.389 V (since 1 Tm²/s = 1 V)

induced current (I) using Ohm's law:

I = ε/R

= 1.389 V / 45 Ω

= 0.03086 A

force (F) from the Lorentz force law, where F = ILB:

F = ILB

= (0.03086 A) * (0.36 m) * (0.65 T)

= 0.00723 N

Finally, we find the power (P) using P = Fv:

P = Fv

= (0.00723 N) * (5.9 m/s)

= 0.04266 W

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Answer:

50,000N

Explanation:

According to Newton's second law of motion;

Net Force = Mass * acceleration

Given

Mass = 5000kg

Let the acceleration = 10m/s²

Net force = 5000 * 10

Net force = 50,000N

Hence the net force acting on the bus is 50000N

Answer:

The difference between (a) and (b) is the deviation caused by the actual pulley not being a perfect solid disk. In (a), we took into account the additional frictional torque and calculated the more accurate moment of inertia. In (b), we made a rough estimate assuming the pulley to be a solid disk, which disregards factors such as the mass distribution and the presence of the axle. The difference between the two values is the deviation caused by these factors.

Answer: mass is neither created nor destroyed

Explanation: For example, the carbon atom in coal becomes carbon dioxide when it is burned. The carbon atom changes from a solid structure to a gas but its mass does not change.

The coefficient of linear expansion for aluminum would be

α = 2.4 x 10⁻⁵.

In the thermal expansion of metals, the ratio of α : β : γ is 1 : 2 : 3.

Given is that an aluminum beam is 10 m long at a temperature of 25 °C. When the temperature of the beam is raised to 75 °C, the bar expands to a final length of 10.012 m.

We can write the change in linear length as -

ΔL = αLΔT

10.012 - 10 = α x 10 x (75 - 25)

0.012 = 10α x 50

0.012/50 = 10α

α = 2.4 x 10⁻⁵

Therefore, the coefficient of linear expansion for aluminum would be

α = 2.4 x 10⁻⁵.

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Answer:

v' = -18 m/s

Explanation:

       \(p_{o} = p_{f} (1)\)

       \(m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)\)

        \(-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)\)

       \(-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)\)

       \(m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)\)