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2 years ago

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a wire carrying 1.5 A passes through a 48 mT magnetic field the wire is peremndicular to the field and makes a quarter circle tu

rn radius 21 cm in the dield regeion. Find the magnitude and direction of the magnetic force

1 answer:

professor190 [17]2 years ago

8 0

Answer:

Explanation:

Wire is in the form of an arc of the size of a quarter of a circle. The distance between its two end

=√2 x 21 cm

= 29.7 cm

= 29.7 x 10⁻² m

Force on the arc in the magnetic field will be same as the straight line meeting its two ends.

Force = BIL

= 48 x10⁻³ x 1.5 x 29.7 x 10⁻²

= 2138.4 x 10⁻⁵ N.

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What is entropy need help ASAP pz

natulia [17]

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2 years ago

Read 2 more answers

A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1

pantera1 [17]

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x = D + (y/2)

1st speaker distance from x = 1.80 + (0.513/2) = 2.0565 m .....1

and

at 2nd speaker distance from x = D - (y/2)

2nd speaker distance from x = 1.80 - (0.513/2) = 1.5435 m .........2

so destructive interference from 1 and 2 we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = ( 0+ 0.5) wavelength

wavelength = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = ( 1 + 0.5) wavelength

wavelength = 0.342 m

frequency 2 = velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

7 0

3 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

3 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

2 years ago

A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of

Citrus2011 [14]

Answer:

a) $(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]$

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

$massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\$

The buoyancy force acting on the balloon is:

$Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]$

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0

2 years ago