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valentina_108 [34]
3 years ago
9

A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A

magnetic force per unit length of 0.130 N/m acts on the conductor in the negative y direction. (a) Determine the magnitude of the magnetic field in the region through which the current passes.
Physics
1 answer:
Jet001 [13]3 years ago
6 0

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

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5p - 14 = 8 p + 4 find p
Digiron [165]

Answer:

Explanation:

5p - 14 = 8p + 4

5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.

-3p = 18. <-- The same thing with the variable itself.

p = -6 <-- Divide both sides by negative 3.

6 0
3 years ago
2. In the activities below indicate if friction is useful or not useful? Explain your answer. i. Writing yes it is useful becaus
Gemiola [76]

Answer:

Please find the answer in the explanation

Explanation:

Friction is a force that opposes motion. One or two of the advantages of friction are break and ability of an object to walk.

Writing yes it is useful because when your writing because friction helps you see what your writing

ii. Rubbing. Yes, it is useful.

friction make it possible for two object to rub each other

iii. Skiing. No. It is not useful because With presence of friction, skiing will not be possible.

iv. Rotating a wheel No. It is not useful because Friction will oppose the rotation of the wheel.

8 0
3 years ago
Severus Snape knows that density of his powder is 3.00 g/cm3 . He also knows he needs 3.00 cm3 of this powder. What mass in gram
harina [27]

Answer:

He requires 1 gram of mass.

Explanation:

The density is defined as:

\rho = \frac{m}{V}  (1)

Where m is the mass and V is the volume.

Then, m can be isolated from equation 1 in order to determine the mass.

m = \rho \cdot V (2)

m = (3.00g/cm^{3})(3.00cm^{3})  

m = 1g  

Hence, he requires 1 gram of mass.

8 0
3 years ago
While strolling downtown on a Saturday afternoon, you stumble across an old car show. As you are walking along an alley toward a
snow_tiger [21]

Answer:

1.44 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v=u+at\\\Rightarrow v=0+a\times 5\\\Rightarrow v=5a

This velocity will be the initial velocity of the car when it passes through the first building

s=ut+\frac{1}{2}at^2\\\Rightarrow 3=5a\times 0.4+\frac{1}{2}\times a\times 0.4^2\\\Rightarrow a=1.44\ m/s^2

The acceleration of the car is 1.44 m/s²

7 0
3 years ago
] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I
Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

\mathtt{=( 0.42\times 9000\times 1055.06) J}

= 3988126.8  J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec =\dfrac{750}{3.99}

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned  = \dfrac{1.1}{100} \times 187.97 \  lb/s

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0
3 years ago
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