An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when the current is 3.70 A?

Answer:1m

Explanation:

2n=0.4m

5n=?

5n×0.4/2n=1m

Answer:

The correct option is;

3. The total mechanical energy is constant as the block moves back and forth

Explanation:

The total mechanical energy is the sum of the potential and kinetic energies of the system

For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant

For a black and spring system, we have total mechanical energy, E = 1/2×K×A².

Where;

K = Constant

A = The amplitude of motion

Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

Answer:

66.86m

Explanation:

Velocity of ball thrown, u = 23.5 m/s

Initial height of the ball above the water, H = 11.5 m

Angle of projection, θ = 39.5°

Vertical components of veloclty = usinθ

Horizontal components of veloclty = ucosθ

The soft ball hits the water after time 't'

Considering the second equation of motion

S = ut + 1/2at^2........ 1

But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:

H = ut +/- 1/2gt^2

H = - 11.5m

U = usinθ

θ = 39.5°

a = -g = -9.8m/s^2

- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2

-11.5m = 23.5(0.6360)t - 4.9t^2

-11.5m = 14.946t - 4.9t^2

4.9t^2 -14.946t-11.5m = 0

Since the ball drifted horizontally

D = (Ucosθ)t

Where θ = 39.5°

U = 23.5m/s t=

Alternatively,

horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s

now how long does it take the ball to raise to a peak and fall to the water.

vertical component of velocity = 23.5 sin 39.5º = 14.947m/s

time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec

peak reached above cliff top is

h = ½gt² = ½(9.8)(1.5252)²

= ½×22.797

= 11.3985m

now the ball has to fall 11.3985+ 11.5 = 22.8985m

time to fall from that height is

t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec

add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869

now haw far does the ball travel horizontally in that time

d = vt = 18.1331 ×3.6869= 66.856m

= 66.86m

Answer:

B.

Explanation:

as there is a '-' sign before the magnification value, it forms a real image.

so the last two options get cancelled.

as the value of magnification is 0.87 ie. lesser than 1, we can say that the image is smaller.

The type of image is formed by a lens if m = -0.87 would be smaller than the object and the object is real

A lens is a transmissive optical tool that employs refraction to focus or disperse a light beam. The power of the lens is expressed in the dioptre which is the reciprocal of the focal length of the lens.

The image formation through the lens is calculated with the help of the lens formula given as follows

1/f = 1/v - 1/u

where f is the focal length of the lens

u is the distance of the object from the lens

v is the distance of the image formed from the

magnification (m) = size of image /size of the object

As given in the problem image is formed by a lens if m = -0.87

Thus, the negative sign represents that the image formed by the object is real, and the magnitude of the magnification is less than 1 this represents that size of the image is smaller than the size of the object.

Learn more about the lens from here

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Answer:

Well if they playing a game like that

Answer:

4.0

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as

Force (F) = mass (m) x acceleration (a)

F = ma

With the above formula, we can obtain th acceleration of the body as follow:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

F = ma

20 = 5 x a

Divide both side by 5

a = 20/5

a = 4 m/s²

Therefore, the value that will complete the last cell in the question above is 4.

Answer:

The distance the car travels is 115500 m in S.I units

Explanation:

Distance d = vt where v = speed of the car and t = time taken to travel

Now v = 99 km/h. We now convert it to S.I units. So

v = 99 km/h = 99 × 1000 m/(1 × 3600 s)

v = 99000 m/3600 s

v = 27.5 m/s

The speed of the car is 27.5 m/s in S.I units

We now convert the time t = 70 minutes to seconds by multiplying it by 60.

So, t = 70 min = 70 × 60 s = 4200 s

The time taken to travel is 4200 s in S.I units

Now the distance, d = vt

d = 27.5 m/s × 4200 s

d = 115500 m

So, the distance the car travels is 115500 m in S.I units

Answer:

Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.

Explanation:

Volume is the amount of space an object takes up, while mass is the amount of matter in an object. ... To find the volume of an irregular sized object, one would use the displacement method for measuring volume and place the object in water and measure the amount of water that is displaced.

Answer:

To measure the volume of an irregularly shaped object, pour some water in a measuring cylinder. Then suspend the irregularly shaped object with a thread. After that , move the object gradually downwards and immerse it in water. The volume of the irregularly shaped object is the difference between the volume of the liquid before and after. After measuring the difference, we come to know about the volume of the irregularly shaped object.

Answer:

I assume its c. Since its talking about testing.

Explanation:

Answer:

The answer is test of durability

Explanation:

Answer; 4.4cm

Explanation: There is a ruler upon which the cylinder start from 1.4cm and reaches 10.2cm

distance traveled =10.2-1.4=8.8

since this cylinder is small so the linear distance can be approximately taked as rotational distance(as in case of point charge) so

2x2πxr =8.8

so the circumference will be 2πr=4.4cm

Answer:

Answer:

1) The density of the object = 2500 kg/m³

2) The density of the oil = 1250 kg/m³

Explanation:

1) The information relating to the question are;

The mass of the object in air = 0.250 kgf

The mass of the object in water = 0.150 kgf

The mass of the object in the oil  = 0.125 kgf

By Archimedes's principle, we have;

The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced

The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf

∴ The weight of the water displaced = 0.1 kgf

Given that the object is completely immersed in the water, we have;

The volume of the water displaced = The volume of the object

The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)

The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³

The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³

The density of the object = 2500 kg/m³

2) Whereby the mass of the object in the oil = 0.125 kgf

The upthrust of the oil = The weight of the oil displaced

The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil

The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf

The weight of the oil displaced = The upthrust of the oil

Given that the volume of the oil displaced = The volume of the oil, we have;

The volume of the oil displaced = 0.0001 m³

The mass of the 0.0001 m³ = 0.125 kg

Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.

The density of the oil = 1250 kg/m³.

Answer:

The answer  is The acceleration is double its original value.

Explanation:

Hope this helps....

Have a nice day!!!!

Answer:

The acceleration is half of its original value

Explanation:

Derived Quantities

Answer:

These are quantities calculated from two or more measurements

Explanation:

They can't me measured directly.

They can only be computed.

They are calculated in PHYSICAL SCIENCE.

hope it helps.

Answer:

c

Explanation:

The marginal cost curve image  has been attached from which we can clearly, indicate that

ATC = average total cost

AFC = average fixed cost

AVC = average variable cost.

From the graph we can indicate that the marginal cost curve

(c) Lies above the AVC curve when the AVC curve slopes downward.

Answer:

1. direction of propagation of sound

2.medium through which sound trsnsmitted

Answer:

They are:

1) change position

2) distract yourself

3) Get fresh air

4) Face the direction you are going.

5) Drink water.

6) Play music.

7) Put your eyes on horizon.  

Explanation:

Hope it helps.

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

The astronomers could know what was happening inside galaxies way back then by the fact that;

they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find

Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.

This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.

The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.

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Answer:

The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.

Answer:

Explanation:

Energy of gamma ray = 2.2233 MeV

Let mass of neutron be n  amu  

mass defect of deuteron = 2.013553212 - ( 1.007 276 467 + n ) u .

in terms of energy this mass defect will be equal to energy of gamma ray

1 amu = 931 MeV

931 [ 2.013553212 - ( 1.007 276 467 + n ) ] =  2.2233

( 1.007 276 467 + n ) -  2.013553212  = .00238807733

n = 1.008664822 amu

so mass of neutron = 1.008664822 amu

Answer:

Rounded to three significant figures:

(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].

Explanation:

Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.  

Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:

The path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].

The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].

Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

The dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.

Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.

For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

[tex]x=4t^3+3t^2-5t+2[/tex]

Where,

x is in meters and t is in sec

We know that,

Velocity,

[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]

(a) i. t = 2 s

[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]

At t = 4 s

[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]

(b) Acceleration,

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]

Pu t = 3 s in above equation

So,

[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]

Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²

Answer:

27.34 (no unit)

Explanation:

26.975*82.33%+29.018*17.67%

=27.34

Answer:

The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.

Answer:

5.3333 sec

Explanation:

initial speed: u = 108km/hr or 30 m/s

final speed: v = 0m/s

distance travelled: s = 80m

time the car took to stop: = t sec

[tex]v^{2} - u^{2}[/tex] = 2as,

a = ([tex]v^{2} - u^{2}[/tex])/2s

a = (0-900)/160

a = -5.625 [tex]ms^{-2}[/tex]

v = u + at,

t = (v - u)/a

t= (0 - 30)/(-5.625)

t = 5.3333 sec

Answer:

The volume of the cavity is 0.013m^3

Explanation:

To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:

Step one:

Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.

Volume of the cylinder = 2.1 / 11.053 =0.19[tex]m^{3}[/tex]

Step two:

From the volume of the cylinder, we can get the radius of the cylinder.

[tex]radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m[/tex]

Step three:

From the cross-sectional area, we can obtain the radius of the cavity.

Let the radius of the cavity be = r, while the radius of the cylinder be = R

CSA of cavity =

[tex]\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m[/tex]

Step Four:

calculate the volume of the cavity using volume =[tex]\pi r^2 \times h[/tex]

Recall that the cavity has the same height as the original cylinder

[tex]volume = \pi \times 0.115^2\times 0.32= 0.013m^3[/tex]

Answer:

Positive

Explanation:

In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive