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Answer:

Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.Nov 28, 2020

Explanation:

Answer:

130n on the 2nd horizontal

Explanation:

I believe you are incorrect. A weathered mountain would appear more jagged.

I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.

If this is incorrect, please, don't refrain to tell me.

Answer:

point 2

Explanation:

Answer:

a sphere of influence mmmmmmmmmmm not actually right

Answer:

Process used for separating grains from the stalks is known as threshing, In this process, stalks are beaten to free the grain seeds.

Answer:

Threshing is extraction of wheat germ from the stalk. In today's usage the combine tractor cuts and threshes the wheat at the same time. Imagine a big lawn mower with a rotating drum inside.

The drum turns and shakes the germ out of the wheat, the seeds falling through small holes onto a conveyor belt one way, the leftover grass dumping out the other way. The grain is poured into a truck driving beside the combine.

In old times, grain had to be beaten out of the grass on a Threshing Floor.

Answer:

is that 4 question each one of them we have to solve

Answer:

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

Explanation:

Let's analyze the situation, as the stars have the same mass and all are at the same distance, the magnitude of the force is constant, even when its direction changes, Let's use Newton's second law

             F = m a

in this case the force is the universal attraction

             F = G M₁ M₂ / r²

they relate it is centripetal

             a = v² / r = w² r

we substitute

             G M² / r² = M w² r

              w² = [tex]\frac{G M}{r^3}[/tex]

angular velocity is related to frequency and period

              w = 2π f = 2π / T

we substitute

              [tex](\frac{2\pi }{T })^2 = \frac{G M}{r^3}[/tex]

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

in this case  r= L

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

Answer:The acceleration due to gravity g is inversely proportional to the square of the radius in the formula g = GM / R^2 where G is the gravitational constant = 6.67 x 10^-11 Nm^2/kg^2, M is the mass of the Earth and R is the radius of the Earth

Explanation:

Answer:

The EM spectrum is generally divided into seven regions, in order of decreasing wavelength and increasing energy and frequency. The common designations are: radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.

Explanation:

HOPE THIS HELPS! ;)

In order to classify the different types of electromagnetic radiation, frequency is necessary.

The EM spectrum have seven different regions in terms of decreasing wavelength. The increase in energy and frequency denotes decreasing wavelength. The parts of EM spectrum are radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.

Thus, in order to classify the different types of electromagnetic radiation, frequency is necessary.

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Answer:

[tex]1290.16\ \text{rad/s}^2[/tex]

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = 734.1 rad/s

[tex]\omega_f[/tex] = Final angular velocity = 0

t = Time = 0.569 seconds

[tex]\alpha[/tex] = Angular acceleration

From the kinematic equations of rotational motion we have

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-734.1}{0.569}\\\Rightarrow \alpha=-1290.16\ \text{rad/s}^2[/tex]

The magnitude of the average angular acceleration of the disk is [tex]1290.16\ \text{rad/s}^2[/tex].

Answer:

Period refers to the time for something to happen and is measured in seconds/cycle. In this case, there are 11 seconds per 33 vibrational cycles. Thus the period is (11 s) / (33 cycles) = 0.33 seconds. We now know that the period is 3.2 seconds and that the frequency is 0.31 Hz.

Answer:

[tex]a=2.5\ m/s^2[/tex]

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]

So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].

Answer:

the velocity of the big fish after the launch is 6 m/s.

Explanation:

Given;

mass of the big fish, m₁ = 5 kg

velocity of the big fish, u₁ = 8 m/s

mass of the small fish, m₂ = 1 kg

velocity of the small fish, u₂ = -4 m/s

Let the final velocity of the big fish after launch = v

Apply the principle of conservation linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

5 x 8   + 1 x (-4) = v(5 + 1)

40 - 4 = 6v

36 = 6v

v = 36/6

v =  6 m/s.

Therefore, the velocity of the big fish after the launch is 6 m/s.

Answer:

a=24.5 b=9.5

Explanation:

The consumer electric bill for 29 days is AED 331.31

Given that, The utility company charges AED 0.077592 per kWh for the electricity plus AED 0.029998 per kWh for the distribution of the electricity

Since, A consumer uses 3098 kWh in 29 days.

The  utility company charges for electricity is,

                                  [tex]=0.077592*3098=240.38[/tex]

The  utility company charges for distribution of the electricity  is,

                                  [tex]=0.029998*3098=92.93[/tex]

So that, The consumer electric bill is,

                         [tex]=240.38+92.93=333.31[/tex]

Hence, the consumer electric bill for 29 days is AED 331.31

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Answer:

b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).

d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).

Explanation:

Two parallel conductors experience attractive force when the current flowing in the conductors are in the same direction.

Also two parallel conductors experience repulsive force when the current flowing in the conductors are in opposite direction.

Therefore, b and d are the correct options.

b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).

d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).

Answer:nutrients

Explanation:

Answer:

pollutants is the right answer.

Answer: perpendicular

Explanation:

Answer:

Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.

Explanation:

Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Explanation:

Answer:

is this it?

Explanation:

λ = h/mv, where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.

Answer:

the velocity of the swimmer at the bottom of the slide is 6.26 m/s

Explanation:

The computation of the velocity of the swimmer at the bottom of the slide is given below:

v = √2gh

= √2 × 9.8 × 2

= 6.26 m/s

Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s

The velocity of the swimmer at the bottom of the slide will be 6.26 m/s.The pace of displacement change  with reference to time is referred to as the velocity

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

The velocity is found as;

[tex]\rm v= \sqrt{2gh} \\\\ \rm v= \sqrt{2\times 9.81 \times 2}\\\\ \rm v= 6.26 \ m/sec[/tex]

Hence velocity of the swimmer at the bottom of the slide will be 6.26 m/s.

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Answer:

a. 1.2×10^-6

b. 0.42×10^9

c. 246.8×10^3

d. 88

Answer:

With a zero signal applied to the Base of the transistor it turns “OFF” acting like an open switch and zero collector current flows. With a positive signal applied to the Base of the transistor it turns “ON” acting like a closed switch and maximum circuit current flows through the device.

Explanation:

please give me brainlist and follow

Answer:

[tex]4.46\times 10^{-4}\ \text{V}[/tex]

Explanation:

B = Magnetic field = [tex]5\times 10^{-5}\ \text{T}[/tex]

r = Radius of rim = 16 m

t = Time = 90 seconds

A = Area of rim = [tex]\pi r^2[/tex]

EMF is given by

[tex]\varepsilon=\dfrac{BA}{t}\\\Rightarrow \varepsilon=\dfrac{5\times 10^{-5}\times \pi\times 16^2}{90}\\\Rightarrow \varepsilon=0.000446=4.46\times 10^{-4}\ \text{V}[/tex]

The magnitude of the induced emf between the hub and the rim is [tex]4.46\times 10^{-4}\ \text{V}[/tex].

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

         W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

         W = K_f                (1)

now let's use conservation of energy

starting point. Highest point A

          Em₀ = U = m g h

Final point. Lowest point B

         Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         mg h = K

to find the height let's use trigonometry

at point A

            cos 35 = x / L

            x = L cos 35

so at the height is

            h = L - L cos 35

            h = L (1-cos 35)

we substitute

           K = m g L (1 -cos 35)

we substitute in equation 1

           W = m g L (1 -cos 35)

let's calculate

           W = 0.500 9.8 0.950 (1 - cos 35)

           W = 0.842 J

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

       [tex]\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)[/tex]  

       [tex]\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)[/tex]

       [tex]3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)[/tex]

       [tex]57.5 rev/sec = 0 + \alpha * 6 s (4)[/tex]

       [tex]\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)[/tex]

       [tex]\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)[/tex]

       [tex]\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)[/tex]

       [tex]\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)[/tex]

       [tex]0 = 57.5 rev/sec + \alpha * 32 s (9)[/tex]

       [tex]\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)[/tex]

        [tex]\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)[/tex]