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3 years ago

13

How large a current would a very long, straight wire have to carry so that the magnetic field 2.20 cm from the wire is equal to

1.00 g (comparable to the earth's northward-pointing magnetic field)?

1 answer:

choli [55]3 years ago

6 0

Biot-Savart Law - Magnetic field around a straight conductor carrying a current I.

B = μ_0 I/2πr

where:
B is strength of magnetic field in T.
μ_0 is permeability of free space = 1.25664 x 10^ -6 T-m/A.
I is current in A.
r is distance from conductor in m.

<span>So I = 2πrB /μ_0</span>

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fuel was consumed at a certain rate of 0.05Kg\s in a rocket engine and ejected as a gas with a speed of4000m\s . Determine the t

ivann1987 [24]

Answer:

Thrust = 200 N

Explanation:

The engine thrust can be found by using the following formula:

$Thrust = mv$

where,

m = mass flow rate of the fuel = 0.05 kg/s

v = velocity of ejected gases = 4000 m/s

Therefore, using the given values in the equation, we get:

$Thrust = (0.05\ kg/s)(4000\ m/s)$

<u>Thrust = 200 N</u>

5 0

3 years ago

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

b

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

c

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

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3 years ago

A cord of negligible mass runs around two massless, frictionless pulleys. A canister with mass m = 20 kg hangs from one pulley.

photoshop1234 [79]

(a) 196 N

The equation of the forces on the side of the cord where the force F is applied is:

$F-T=0$ (1)

where T is the tension in the cord.

On the other side of the cord, the equation of the forces on the canister is

$T-mg = ma$ (2)

where

m = 20 kg is the mass of the canister

$g=9.8 m/s^2$ is the acceleration of gravity

a is the acceleration

From (1),

$T=F$

Substituting into (2),

$F-mg = ma\\F=m(g+a)$

We want the canister to move at constant speed, so

a = 0

And therefore:

$F=mg=(20)(9.8)=196 N$

b) 2.0 cm

The cord is inextensible, this means that the acceleration of its parts are the same. Therefore, the acceleration of the free end must be the same as the acceleration of the canister: and this means that the two parts also cover the same distance in the same time.

Therefore, the free end of the cord must be moved exactly the same as the canister, by 2.0 cm.

c) 3.92 J, the same

The work done by the tension in the cord is

$W_T = T d$

where

T is the tension

d = 2.0 cm = 0.02 m is the displacement

As we said in part (a), the tension in the cord is equal to the force applied to the free end:

T = F

So

T = 196 N

Therefore, the work done by the tension is

$W=(196)(0.02)=3.92 J$

And since the force applied (F) is the same, then the work done by you when pulling the cord is exactly the same.

(d) -3.92 J

The weight of the canister is

$F_g = mg =(20 kg)(9.8 m/s^2)=196 N$

However, the direction of the force of gravity is opposite to the displacement. Therefore, the work done by gravity is negative:

$W_g = - F_g d$

And substituting,

$W_g=-(196)(0.02)=-3.92 J$

(e) Zero

The net work done on the canister can be simply calculated by adding the work done by the tension in the cord and the weight of the canister:

$W=W_T+W_g = 3.92 + (-3.92 ) = 0$

This is in agreement with the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. In this situation, the canister is moving at constant speed, so its kinetic energy is not change: therefore,

$\Delta K = 0$ (change in kinetic energy = 0)

and so, the work done on it is also zero.

(f) The pulley system changes the direction of the force applied

This is a simple pulley system, which means that the system does not multiply the force applied in input. In fact, the mechanical advantage of the system is

$MA=\frac{F_{out}}{F_{in}}$

where:

$F_{out}$ is the output force, which is the weight of the canister

$F_{in}$ is the force in input, which is F

So, the mechanical advantage is 1:

$MA=\frac{196 N}{196 N}=1$

From a point of view of energy, therefore, there is no advantage in this system.

However, the advantage offered by the pulley system concerns the direction of the force: in fact, it changes the direction of the applied force (which is F, downward) into the tension of the cord (which is upward on the canister).

6 0

4 years ago

A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the

Gelneren [198K]

Answer:

Option B. 2.8 s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 27 m/s

Angle of projection (θ) = 30

Acceleration due to gravity (g) = 9.8 m/s²

Time of flight (T) =?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = 2 × 27 × Sine 30 / 9.8

T = 2 × 27 × 0.5 / 9.8

T = 27 / 9.8

T = 2.8 s

Therefore, time of flight of the ball is 2.8 s

7 0

3 years ago

How can astrophysicists tell whether a star is receding from or approaching earth?

MrMuchimi

Answer:

Doppler shift of the starlight

Explanation:

To predict the movement of a star, we compare the spectra of elements found in star (H, He Na etc.), first spectra which are obtained from star and second spectra from laboratory. If spectral lines of the spectra obtained from star, are shifting towards red end (called red shift) then star is going away from earth and if shifting is towards blue (called blue shift), then star is approaching the earth. This is Doppler's shift.

3 0

4 years ago