A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits the water below. What was the pelican’s initial speed? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s. If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

Answer:

The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.

Explanation:

Given that,

Acceleration = 9.5 m/s²

Diameter = 283 m

We need to calculate the frequency of the rotational motion for the wheel to produce this effect

Using formula of rotational frequency

[tex]a= r\omega^2[/tex]

[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]

Where, r = radius

a = acceleration

[tex]\omega[/tex] = rotational frequency

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{9.5\times2}{283}}[/tex]

[tex]\omega=0.259\ rad/s[/tex]

The frequency in rev/min

[tex]\omega=0.259\times\dfrac{60}{2\pi}[/tex]

[tex]\omega=2.473\ rev/min[/tex]

Hence, The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.

Answer:

B

Explanation:

cause it makes the most sence

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

Answer:

0.45 V and 0.9 V.

Explanation:

To test a PNP transistor with an ohmmeter,

Plug the positive lead from the multimeter to the transistor EMITTER (E). Plug the negative meter into the transistor BASE (B). If you are using a PNP resistor you must watch OL that is over limit, the voltage decrease will indicate between 0.45V and 0.9V if you are measuring it.

Testing a PNP transistor with an ohmmeter should yield high resistance values when the base is negative relative to the emitter or collector, and low resistance values when the base is positive relative to the emitter or collector.

When testing a PNP transistor with an ohmmeter, the high or low resistance values that are expected for a good transistor are as follows: When you measure between the base and the emitter or collector, you should see a high resistance value (in the range of megaohms) if the base is negative with respect to the emitter or collector. Conversely, you should see a low resistance value (in the range of a few hundred Ohms) when the base is positive with respect to the emitter or collector.

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Answer:

Q = 12.5 kJ

Explanation:

The expression to use to calculate Heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as it was stated in the comments, the question is incomplete, and here is the missing part:

Given:

2A + B  A2B (1)

H° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

H° = 35.0 kJ/mol

With these two reactions, we can calculate the heat.

Now, with the above two reactions, we need to get the general reaction (The one the question is giving), so, let's use (1) and (2) and do the sum of them:

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

Now, we sum both equations, we can see that one A2B cancels out with one A2B from equation 2, so, the equation gives:

2A + B + 2A2B -------> 2AB + A2

And the enthalpy, it's just summed:

H°3 = -25 + 35 = 10 kJ/mol

Now with this value we can calculate heat:

Q = 10 * 2.5 = 25 kJ

However, in the reaction we have 2A, so it's not 1:1 mole ratio, but instead is 1:2, so this result we have to divide it between 2 so:

Q = 25 / 2 = 12.5 kJ

The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out.

To calculate the heat required for the reaction, we need to know the enthalpy change (ΔH°) for the reaction 2a + b → a2b. However, the enthalpy change for this reaction is not provided in the question. Assuming that the enthalpy change for this reaction is known, we can proceed with the calculation.

 Let's denote the enthalpy change for the reaction 2a + b → a2b as ΔH°. The reaction of interest is:

[tex]\[ 2a + b + a_2b \rightarrow 2ab + a_2 \][/tex]

 This reaction can be broken down into two parts:

1. The reaction of 2 moles of a with 1 mole of b to form 1 mole of a2b:

with an enthalpy change of H°.

[tex]\[ 2a + b \rightarrow a_2b \][/tex]

2. The reaction of 1 mole of a2b to form 2 moles of ab:

[tex]\[ a_2b \rightarrow 2ab \][/tex]

Since the reaction of a2b to form 2ab does not involve any additional reactants, it can be considered as the reverse of the first part, but with twice the amount of ab produced. Therefore, the enthalpy change for this part would be -H° (since the reverse reaction has the opposite sign of the forward reaction), and for 1 mole of a2b reacting, it would be -2ΔH° (because 2 moles of ab are formed).

Now, we are given 2.50 moles of a, and we assume there is excess b and a2b. The reaction will proceed until all of a is consumed. Since the reaction stoichiometry is 2 moles of a to 1 mole of a2b, 1.25 moles of a2b will be required to react with the 2.50 moles of a (since 2.50 moles of a require 1.25 moles of a2b).

The overall enthalpy change for the reaction will be the sum of the enthalpy changes for the two parts:

 1. For the reaction of 2.50 moles of a with b to form a2b:

[tex]\[ \Delta H_1 = 2.50 \text{ moles} \times \Delta H^\circ \][/tex]

2. For the reaction of 1.25 moles of a2b to form ab:

[tex]\[ \Delta H_2 = 1.25 \text{ moles} \times (-2 \Delta H^\circ) \][/tex]

The total enthalpy change (ΔH_total) is the sum of ΔH_1 and ΔH_2:

[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 \][/tex][tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2 \times 1.25 \Delta H^\circ \][/tex]

[tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2.50 \Delta H^\circ \][/tex]

[tex]\[ \Delta H_{\text{total}} = 0 \][/tex]

 Surprisingly, the total enthalpy change for the reaction is zero. This is because the amount of heat released in the formation of a2b is exactly equal to the amount of heat absorbed in the formation of ab from a2b. Therefore, no additional heat is required for the overall reaction to proceed.

 The final answer is:  [tex]\[ \boxed{0} \][/tex]

The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out. This conclusion is based on the assumption that the enthalpy change for the reaction 2a + b → a2b is known and that the reaction proceeds as written. If the enthalpy change for the formation of a2b is not known, then it would be necessary to look up the standard enthalpy of formation for a2b to perform the calculation.

Answer:

20 metres

Explanation:

Speed = distance ÷ time

[tex]s = \frac{d}{t} [/tex]

If we substitute the values:

[tex]4 = \frac{d}{5} [/tex]

[tex]20 = d[/tex]

Answer: 20m

Explanation:

Speed = Distance/time taken

Speed = 4.00m/s

Time taken = 5.00s

Distance = D = ?

We insert the values in the formula

4.00m/s = D/5.00s

Multiply through by 5

D = 20m

Answer:

E=147898.01J

Explanation:

A 5.0-kg cannonball is fired over level ground with a velocity of 2.00 ⨯ 102 m/s at an angle of 25° above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip, find the change in the thermal energy of the cannonball and air

firstly , we look for the time of flight it takes to make the projectile path

T=2Usin∅/g

take g= 9.81m/s

T=2*200sin25/(9.81)

T=17.23Secs

energy is force *distance

E=f*d

f=m*g

f=5*9.81

f=49.05N

s=distance

s=(v+u)T/2

s=(150+200)17.23/2

s=3015.25m

49.05N*3015.25m

E=147898.01J

Answer: 0.471 m/s

Explanation:

We are given the followin equation:

[tex]Re=\frac{D v \rho}{\eta}[/tex] (1)

Where:

[tex]Re[/tex] is the Reynolds Number, which is adimensional and indicates if the flow is laminar or turbulent

When [tex]Re<2100[/tex] we have a laminar flow

When [tex]Re>4000[/tex] we have a turbulent flow

When [tex]2100<Re<4000[/tex] the flow is in the transition region

[tex]D=2R[/tex] is the diameter of the pipe. If the pipe ha a radius [tex]R=8(10)^{-3} m[/tex] its diameter is [tex]D=2(8(10)^{-3} m)=0.016 m[/tex]

[tex]v[/tex] is the average speed of the fluid

[tex]\rho=1060 kg/m^{3}[/tex] is the density of the fluid

[tex]\eta=4(10)^{-3} Pa.s[/tex] is the viscosity of the fluid

Isolating [tex]v[/tex]:

[tex]v=\frac{Re \eta}{D \rho}[/tex] (2)

Solving for [tex]Re=2000[/tex]

[tex]v=\frac{(2000)(4(10)^{-3} Pa.s)}{(0.016 m)(1060 kg/m^{3})}[/tex] (3)

Finally:

[tex]v=0.471 m/s[/tex]

Answer:

Rate of child doing work on box = μmgv (Unit is Watt)

Explanation:

Rate of child doing work on box = Work done / time = Power  

Power = Horizontal force x Velocity

We are aware that the Velocity in this case is v.

As the object is moving with constant velocity, the acceleration would be zero and the applied horizontal force will be equal to friction force. So in our case,  

Horizontal force = friction force

We know that the coefficient of friction is the ratio of friction force to Normal force,

μ = friction force / Normal force

Normal Force = mg,  where m is the mass and g is the gravitational acceleration

Friction force = μ x Normal Force

Friction force = μmg

Power = μmgv (Unit of power is Watt)

Answer:

P = μ*mg*v

Explanation:

A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ.  The rate at which the child works is calculated as shown below:

mass of the box = m; coefficient of friction is μ; speed = v.

In order to push the box, the child must exert a force equal to or more than the frictional force.

force = coefficient of friction*weight of the box

f = μ*mg

In addition, to calculate the rate of work (i.e. power). We have:

Power = force*velocity (or speed)

Therefore:

P = μ*mg*v

Answer:

V(h) = Eh

Explanation:

I will assume that the capacitor is a parallel-plate capacitor.

By Gauss' Law, electric field inside the capacitor is

[tex]E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}[/tex]

The relation between electric field and potential is

[tex]V_{ab} = -\int\limits^b_a {\vec{E}(h)} \, d\vec{h}  = \int\limits^h_0 {\frac{Q}{\epsilon_0 A}} \, dh \\V(h) - V(0) = V(h) - 0 = Eh\\V(h) = Eh[/tex]

The important thing in this question is that the electric field inside the parallel plate is constant. So, the potential is also constant and proportional to the distance, h.

The electric potential V(h) inside a capacitor as a function of height h, with zero potential at the bottom plate, is V(h) = Eh, using the relationship between the electric field E and potential V where E is constant.

To calculate the electric potential V(h) inside a capacitor as a function of height h, with the potential at the bottom plate taken to be zero, you can use a relation between the electric field E and the potential V. Given that E = V/d, where d is the separation between the plates, and that the electric field E is uniform, we have the relationship E = -dV/dh (the negative sign indicates the direction of the potential decrease). Integrating this equation from 0 to h, where V(0) = 0, gives us V(h) = -Eh. However, we can ignore the negative sign because we are interested in magnitude.

So, the final expression for V(h) inside the capacitor, in terms of the electric field E and the height h above the bottom plate, is:

V(h) = Eh

Answer:

The time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours

Explanation:

Electric Charge: This is defined as the product of electric current to time in a an electric circuit. The S.I unit of charge is Coulombs (C).

Mathematically, it is represented as

Q = it.......................... Equation 1

Where Q =quantity of charge, I = current, t = time

making t the subject of formula in equation 1,

t = Q/I ....................... Equation 2

Given: Q = 3.5 C, I = 5 mA.

Conversion: We convert from mA to A

I.e 5 mA = (5 × 10⁻³) A = 0.005 A.

Substituting these values into equation 2

t = 3.5/0.005

t = 700 seconds  

or

(700/60) minutes = 11.67 minutes

or

(700/3600) hours = 0.194 hours.

Therefore the time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours

700 second will take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes.

What information do we have?

Charges of partical = 3.5 C

Current = 5 mA = (5 × 10⁻³) A

Charge = (Curernt)(time)

Q = iT

T = Q / i

Time taken = 3.5 / (5 × 10⁻³)

Time taken = 700 seconds or 11.67 minutes

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Answer:odd thousands plus 500 feet.

Explanation:

On a magnetic course of zero through 179 degrees, select an odd thousand foot cruising altitude plus 500 feet, such as 3,500, 5,500, up to and including 17,500. Even and odd thousands are reserved for those aircraft on an active instrument flight plan. Even thousands plus 500 feet are for aircraft flying between 180 and 359 degrees.

Answer:

27.5 days

0.92 month

Explanation:

[tex]r[/tex] = radius of the orbit of moon around the earth = [tex]3.85\times10^{8} m[/tex]

[tex]M[/tex] = Mass of earth = [tex]5.98\times10^{24} m[/tex]

[tex]T[/tex] = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

[tex]T^{2} = \frac{4\pi ^{2} r^{3}  }{GM}[/tex]

inserting the values, we get

[tex]T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3}  }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec[/tex]

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

[tex]T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days[/tex]

1 month = 30 days

[tex]T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month[/tex]

Final answer:

The period for the moon's motion around the earth is approximately 0.59 days, which is much shorter than the length of a month.

Explanation:

To find the period for the moon's motion around the earth, we can use Kepler's third law. According to Kepler's third law, the square of the period of a planet's orbit is directly proportional to the cube of its average distance from the center of the orbit.

We are given that the moon orbits the earth at a distance of 3.85 x 10^8 m. We can use this information to calculate the period as follows:

Hence, the period for the moon's motion around the earth is approximately 0.59 days. This is much shorter than the length of a month, which is about 30 days. Therefore, the moon completes multiple orbits around the earth in one month.

Answer:

It takes 10.5 minutes to kill all the bacteria.

Only 1 cell would remain after 9 minutes.

Explanation:

It will take 1.5 minutes to kill 90% of the cells. So, after 1.5 minutes, only 10% would remain. After 3 minutes, only 1% remain. So, to figure out how long it would take to kill a million cells, we have to multiply 1 million by 0.1 repeatedly until the final value is less than 1 that is because when the value is less than 1, it means there are no more bacteria.

So:  

[tex]10^6 \times (0.1)^7[/tex] = 0.1  

So, you need 10.5 minutes of killing to kill one million cells.

Time taken=  7 x 1.5 minutes = 10.5 minutes.  

After 9 minutes you would have:  

[tex]10^{6} \times (0.1)^{6}[/tex] = 1 cell left

The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells.

The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. Since the DRT value represents the time it takes to kill 90% of the cells, you can calculate the time to kill 100% of the cells by dividing the DRT value by 90 and then multiplying it by 100. In this case, it would be as follows:

T = (1.5 minutes / 90) * 100 = 1.67 minutes

If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells. As a result, some cells would still be alive.

Answer:

Generally, 40 to 50 degrees

Explanation:

About the heat-up over time, whether the windows of a vehicle are locked or partially open makes very little difference. In both situations, in an internal temperature of a vehicle, even at an outside temperature of only 72 ° F, it may exceed approximately 40 ° F within one hour. This happens mainly due to the greenhouse effect that is the heat inside the car is trapped and not allowed to escape. Thus, raising the temperature of the vehicle.

Final answer:

The greenhouse effect causes the temperature inside a parked car to be much higher than outside, sometimes 20 to 30 degrees higher, due to sunlight being trapped as heat.

Explanation:

The temperature inside a vehicle can rise significantly higher than the outside temperature due to a phenomenon known as the greenhouse effect. When a car is parked in the sun with windows closed, the sunlight passes through the glass and is absorbed by interior surfaces such as dashboard and seats. These surfaces then emit infrared radiation, which cannot escape back through the glass, and consequently, the trapped heat raises the temperature of the air inside the car. This effect can cause interior temperatures to be much higher—often 20 to 30 degrees higher—than the external air temperature.

Answer:

its a A. Transverse wave

Answer:

Because of heavy mass

Explanation:

When force acts on a body it tends to accelerate the body. The acceleration produced in the body depends on two things:

1). Magnitude of force

2). Mass of the body

F= ma

⇒ a = F/m  

As the force exerted on earth and another object are the equal in magnitude but opposite in direction. This forces will accelerate the object toward the earth but can't accelerate the earth as earth has very high mass.

a = F/m

This force tends to accelerate the earth but but due to earth's inertia the earth does not accelerate.

Answer:

In computing the volume of a cube,

Maximum possible error = +/-1350cm³

Relative error = 0.05

Percentage error = 5%

In computing the surface area of a cube,

Maximum possible error = +/-180cm²

Relative error = 0.0333

Percentage error = 3.33%

Explanation:

A cube is a three dimensional solid object with six (6) faces, twelve (12) edges and eight(8) vertices.

The volume of a cube = x³

Where x= length of the edge of a cube

X = 30cm +/- 0.5cm

Differentiate V with respect to x (V = Volume of a cube)

dV/dx = 3 x²

dV = 3 x² . dx

dV= 3 × 30² × (+/-0.5)

= 2700(+/-0.5)

= +/-1350cm³

Maximum possible error =

+/- 1350cm³

Relative error = Maximum error /surface area

= ΔV/V

Recall that V = x³

V= (30)³

A = 27000cm³

Substitute the values for and V into the formula for Relative error

Relative error = 1350 / 270000

Relative error = 0.05

% error = Relative error × 100

= 0.05× 100

= 5%

Surface Area of a cube = 6x²

A = 6x²

Differentiate A with respect to x

dA/dx= 12x

dA = 12x . dx

dA= 12 × 30 (0.5)

= +/- 180cm²

Maximum possible error =

+/- 180cm²

Relative error = Maximum error / total area

= dA/dx

Recall that A = 6x²

A = 6(30)²

A = 5400cm²

Substitute the values for and A into the formula for Relative error

Relative error = 180/ 5400

Relative error = 0.0333(4 decimal place)

% error = Relative error × 100

= 0.0333 × 100

= 3.33%

Final answer:

Use of differentials to estimate maximum and relative errors in volume and surface area calculations for a cube.

Explanation:

Differentials for Cube:

Surface Area:

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹

[tex]d_{ab}[/tex] = distance traveled from station A to station B

[tex]t_{ab}[/tex] = time of travel between station A to station B

we know that

[tex]Time = \frac{distance}{speed}[/tex]

[tex]t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}[/tex]

[tex]d_{bc}[/tex] = distance traveled from station B to station C

[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹

[tex]t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}[/tex]

Total distance traveled is given as

[tex]d = d_{ab} + d_{bc}[/tex]

Total time of travel is given as

[tex]t = t_{ab} + t_{bc}[/tex]

Average speed is given as

[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }[/tex]

Given that :

[tex]d_{ab} = 4 d_{bc}[/tex]

So

[tex]v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}[/tex]

2)

[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹

[tex]t_{ab}[/tex] = time of travel between station A to station B

[tex]d_{ab}[/tex] = distance traveled from station A to station B

we know that

[tex]distance = (speed) (time)[/tex]

[tex]d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}[/tex]

[tex]d_{bc}[/tex] = distance traveled from station B to station C

[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹

[tex]t_{bc}[/tex] = time of travel for train from station B to station C

we know that

[tex]distance = (speed) (time)[/tex]

[tex]d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}[/tex]

Total distance traveled is given as

[tex]d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}[/tex]

Total time of travel is given as

[tex]t = t_{ab} + t_{bc}[/tex]

Average speed is given as

[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}[/tex]

Given that :

[tex]t_{ab} = 3 t_{bc}[/tex]

So

[tex]v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}[/tex]

Answer:

[tex]0.54^{\circ}[/tex]

3.99322032 mm

Explanation:

n = Lines per mm = 125

Seperation between slits is given by

[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{125}\\\Rightarrow d=0.008\ mm[/tex]

[tex]\lambda_1[/tex] = 498 nm

[tex]\lambda_2[/tex] = 569 nm

We have the expression

[tex]dsin\theta_1=m\lambda_1[/tex]

For first maximum m = 1

[tex]\theta_1=sin^{-1}\dfrac{m\lambda_1}{d}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1\times 498\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_1=3.57^{\circ}[/tex]

[tex]\theta_2=sin^{-1}\dfrac{m\lambda_2}{d}\\\Rightarrow \theta_2=sin^{-1}\dfrac{1\times 569\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_2=4.08^{\circ}[/tex]

Angular separation is given by

[tex]\Delta \theta=\theta_2-\theta_1\\\Rightarrow \Delta \theta=4.08-3.57\\\Rightarrow \Delta \theta=0.54^{\circ}[/tex]

Angular separation is [tex]0.54^{\circ}[/tex]

Now

[tex]\lambda_1[/tex] = 589 nm

[tex]\lambda_2[/tex] = 589.59 nm

[tex]\Delta \lambda=\lambda_2-\lambda_1\\\Rightarrow \Delta \lambda=589.59-589\\\Rightarrow \Delta \lambda=0.59]\ nm[/tex]

We have the relation

[tex]\dfrac{\lambda}{\Delta \lambda}=mN\\\Rightarrow N=\dfrac{\lambda}{m\Delta \lambda}\\\Rightarrow N=\dfrac{589}{2\times 0.59}\\\Rightarrow N=499.15254[/tex]

Width is given by

[tex]w=\dfrac{N}{n}\\\Rightarrow w=\dfrac{499.15254}{125}\\\Rightarrow w=3.99322032\ mm[/tex]

The width is 3.99322032 mm

The angular separation  \theta of the first maxima of these spectral lines generated by this diffraction grating is 0.54°

The width which this grating needs to be to allow you to resolve the two lines 589.00 and 589.59 nanometers is 3.99322032 mm

n = Lines per mm

= 125

The Separation between slits is given by:

d= 1/n

d= 1/125

= 0.008mm.

Where

line 1 = 498nm

line 2 = 569nm

The first maximum m= 1 will be:

θ1=  3.57°

θ2= 4.08°

The angular separation would be:

θ2- θ1= 0.54°.

Now, to find the width is:

w= N/n

= 499.15254/125

= 3.99322032 mm.

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Answer:

Explanation:

(a)mass [tex]m= 2.5 kg [/tex]

height [tex]h=6 m[/tex]

work required to raise

[tex]W=mgh[/tex]

[tex]W=2.5\times 9.8\times 6=147 J[/tex]

(b)Force [tex]F=20.8 N[/tex]

mass of cart [tex]m=3 kg[/tex]

length of track [tex]s=0.636 m[/tex]

[tex]Work\ done=F\cdot s[/tex]

Work done[tex]=20.8\cdot 0.636=13.22 J[/tex]

(c)mass of eddy [tex]m_e=65 kg[/tex]

height climbed [tex]h=1.6 m[/tex]

time [tex]t=1.2 s[/tex]

Energy required [tex]E=mgh=65\times 9.8\times 1.6=1019.2 J[/tex]

[tex]power=\frac{E}{t}=\frac{1019.2}{1.2}=849.33 W[/tex]

The work done in lifting the 2.5 kg object is 147 J, in moving the 3-kg cart is 13.2 J, and the power used by Eddy in climbing stairs is 851.7 W.

To solve these problems, we need to apply principles of physics, specifically related to work, energy, and power. For the first question, we use the concept of gravitational potential energy, which is calculated by multiplying together the object's mass, the acceleration due to gravity (~9.81 m/s² on Earth), and the height to which the object is lifted. Thus for the 2.5-kg object, the work done or energy required to lift it to a height of 6.0 meters is: W = m * g * h = 2.5 kg * 9.8 m/s² * 6.0 m = 147 J.

Next, for the 3-kg cart, since the cart moves at constant speed, we can say the work done is equal to the change in potential energy. Therefore, the work done is W = m * g * h = 3.0 kg * 9.8 m/s² * 0.450 m = 13.2 J.

Fianlly, for Eddy's case, power is defined as the work done per unit time. If Eddy lifts his own mass to the height of 1.6 m, the work done (again considering as change in gravitational potential energy) is W = m * g * h = 65 kg * 9.8 m/s² * 1.6 m = 1022 J. Given that he does this work in 1.2 seconds, the power expended would be P = W / t = 1022 J / 1.2 s = 851.7 W.

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Answer:

C. all above is true.

Explanation:

Energy releasing reactions are exothermic. Total energy of products ( [tex]  N_2 H_2[/tex]  ) is less than the total energy of reactants ( [tex]  H_2 + N_2 [/tex] ) gives negative enthalpy change.

hint: prefix exo- means "outside, external".

Answer:4 Hz

Explanation:

Speed of wave [tex]v=5 m/s[/tex]

crest to crest distance [tex]\lambda =2 m[/tex]

velocity of observer [tex]v_0=3 m/s[/tex]

actual frequency [tex]f=\frac{velocity}{\lambda }[/tex]

[tex]f=\frac{5}{2}=2.5 Hz[/tex]

Apparent frequency [tex]f'=f(\frac{v+v_0}{v})[/tex]

[tex]f'=2.5\times \frac{5+3}{5}[/tex]

[tex]f'=4 Hz[/tex]      

The frequency of these waves is 4hz.

The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m. Alo, A person in a motorboat head directly toward S at 3.0 m/s.

The actual frequency is [tex]= 5 \div 2[/tex] = 2.5Hz

Now

Apparent frequency is

[tex]= 2.5 \times (5 + 3) \div 5[/tex]

= 4hz

Hence, the frequency of these waves is 4hz.

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Answer:

Ф1=295.96Nm^2/C

 Ф2=469.73Nm^2/C

Explanation:

The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0

 we have two surfaces where we know angle of surface 1 lets call it

s1= 34.

Therefore to find s2

s2 (the angle from surface 2) we have that

s2=180-(90+s1),

so s2=180-(90+34),

thus s2=56 degrees.

Flux equation reads as Φ=ΕΑ,

where Φ is the flux,

E is the electric field and

A is the surface area.

So with respect to the angles and the figure provided,

we have Φ=EAcos(s).

So we can solve further by writing

. For Surface 1 we have

Φ1=EAcos(s1)=210 x 1.7 x cos 34,

so Φ1=295.96,

approximately to an whole number

Φ1=296 Nm^2/c.

Similarly for Φ2, we have

Φ2=EAcos(s2)

=210 x 4 x cos34=469.7,

thus Φ2=469.7Nm^2/c.

Answer:

F= 15.6 N,   a= 33.2 m/s^2

Explanation:

mass= m = 0.47 kg

spring constant= k = 130N/m

spring compression = x = 0.12 m

a).

force on the mass= F = k*x

F = 130 * 0.12 N

F= 15.6 N

b).

Acceleration of mass= a=?

F= ma

a=F / m

a= 15.6/ 0.47 m/s^2

a= 33.2 m/s^2

Final answer:

The force on the mass when the spring is released is 15.6 N, and the acceleration of the mass at that instant is approximately 33.19 m/s^2, calculated using Hooke's Law and Newton's second law, respectively.

Explanation:

Understanding Spring Force and Acceleration

To find the force on the mass at the instant the spring is released (in part A), we use Hooke's Law, which states that the force exerted by a spring (F) is equal to the negative spring constant (k) times the displacement from equilibrium (x), so F = -kx. Here, k = 130 N/m and x = 0.12 m, so the force is F = 130 N/m * 0.12 m = 15.6 N. The negative sign indicates that the force is in the opposite direction of the displacement.

To find the acceleration of the mass at the instant the spring is released (in part B), we apply Newton's second law, which relates force (F), mass (m), and acceleration (a) as F = ma. Rearranging for acceleration, we get a = F/m. Substituting the values, we have a = 15.6 N / 0.47 kg = approximately 33.19 m/s2.

Answer:There are three types of radiation

Alpha, Beta and Gamma radiation

Explanation: In an electric field produced by two parallel charged plates alpha particle would be deflected toward a - plate following a parabolic path, beta rays toward a +plate following a parabolic path and gamma radiation either - or + source.

Answer:

option D

Explanation:

given,

coefficient of friction between wall and tire = µ

speed of motorcycle = s

friction force = f = μ N

where normal force will be equal to centripetal force

[tex]N = \dfrac{mv^2}{r}[/tex]

for motorcycle to not to slip weight should equal to the centripetal force

 now,

[tex]m g =\mu \dfrac{mv^2}{r}[/tex]

[tex]\mu =\dfrac{rg}{s^2}[/tex]

where "rg" is constant

[tex]\mu\ \alpha \ \dfrac{1}{s^2}[/tex]

[tex]\mu\ \alpha \ s^{-2}[/tex]

Hence, the correct answer is option D

Answer:

Δ v =  125 m/s

Explanation:

given,

mass of space craft = 435 Kg

thrust = 0.09 N

time = 1 week

       = 7 x 24  x 60 x 60 s

change in speed of craft = ?

Assuming no external force is exerted on the space craft

now,

[tex]T= m_s a[/tex]

[tex]a=\dfrac{T}{m_s}[/tex]

[tex]a =\dfrac{0.09}{435}[/tex]

a = 2.068 x 10⁻⁴ m/s²

using equation of motion

Δ v = a t

Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60

Δ v =  125 m/s

Answer:

MOTOR

Explanation:

The device which changes electrical energy into mechanical energy is known as motor

Loops of wire in a magnetic field make up motors. The magnetic field produces a torque on the loops when current flows through them, turning a shaft as a result.

Electrical energy is transformed into mechanical energy by electric motors. When we turn on the fan, for instance, the electric motor begins to transform the electrical energy into mechanical energy.

The fan blades then begin whirling as a result of the mechanical energy, giving them the capacity to perform work.

On either side of the controller, there is a motor. This engine has an uneven weight linked to it.

This indicates that one side of it is heavier than the other. The imbalance of the weight causes the controller to vibrate when the motor turns.

Therefore, Mechanical work is created as a result of the conversion of electrical energy.

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Answer:

a)

6.33 x 10⁸ N/C

Direction : Towards negative charge.

b)

1.11125 x 10²⁰ m/s²

Direction : Towards positive charge.

Explanation:

a)

[tex]Q_{1}[/tex] = magnitude of negative charge = 25 x 10⁻⁶ C

[tex]Q_{2}[/tex] = magnitude of positive charge = 50 x 10⁻⁶ C

[tex]r_{1}[/tex] = distance of negative charge from point P = 0.02 m

[tex]r_{2}[/tex] = distance of positive charge from point P = 0.08 m

Magnitude of electric field at P due to negative charge is given as

[tex]E_{1} = \frac{kQ_{1}}{r_{1}^{2} } = \frac{(9\times10^{9})(25\times10^{-6})}{0.02^{2} } = 5.625\times10^{8} N/C[/tex]

Magnitude of electric field at P due to positive charge is given as

[tex]E_{2} = \frac{kQ_{2}}{r_{2}^{2} } = \frac{(9\times10^{9})(50\times10^{-6})}{0.08^{2} } = 0.703125\times10^{8} N/C[/tex]

Net electric field at P is given as

[tex]E = E_{1} + E_{2}\\E = 5.625\times10^{8} + 0.703125\times10^{8} \\E = 6.33\times10^{8} N/C[/tex]

Direction:

Towards the negative charge.

b)

[tex]m[/tex] = mass of the electron placed at P = 9.31 x 10⁻³¹ C

[tex]Q_{1}[/tex] = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

Acceleration of the electron due to the electric field at P is given as

[tex]a = \frac{qE}{m}\\ a = \frac{(1.6\times10^{-19})(6.33\times10^{8})}{(9.11\times10^{-31})}\\a = 1.11125\times10^{20} ms^{-2}[/tex]

Direction: Towards the positive charge Since a negative charge experience electric force in opposite direction of the electric field.

Answer:

Explanation:

qA = - 25 x 10^-6 C

qB = 50 x 10^-6 C

AP = 2 cm

BP = 8 cm

(a)

Electric field at P due to the charge at A

[tex]E_{A}=\frac{kq_{A}}{AP^{2}}[/tex]

[tex]E_{A}=\frac{9\times 10^{9}\times 25\times 10^{-6}}{0.02^{2}}[/tex]

EA = 5.625 x 10^8 N/C

Electric field at P due to the charge at B

[tex]E_{B}=\frac{kq_{B}}{BP^{2}}[/tex]

[tex]E_{A}=\frac{9\times 10^{9}\times 50\times 10^{-6}}{0.08^{2}}[/tex]

EB = 0.70 x 10^8 N/C

The resultant electric field at P due to both the charges is

E = EA+ EB = (5.625 + 0.7) x 10^8

E = 6.325 x 10^8 N/C towards left

(b)  mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of electron

Force on electron, F = charge of electron x electric field

F = q x E

[tex]a = \frac{qE}{m}[/tex]

[tex]a = \frac{1.6\times 10^{-19}\times 6.325\times 10^{8}}{9.1\times 10^{-31}}[/tex]

a = 1.11 x 10^20 m/s^2