Ask question

Ask question

All categories

stepladder [879]

3 years ago

14

An equilateral triangle 8.0 cm on a side is in a 3.5 mT uniform magnetic field. The magnetic flux through the triangle is 6.0 μW

b. What is the angle between the magnetic field and an axis perpendicular to the plane of the triangle?

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer:

43.80°

Explanation:

The side of the equilateral triangle is given as 8 cm

The area of the equilateral triangle $A=\frac{\sqrt{3}}{4}a^2=\frac{1.732\times 8^2}{4}=27.712cm^2=27.712\times 10^{-4}m^2$

Magnetic field $B=3.5mT=3.5\times 10^{-3}T$

Magnetic flux $\Phi =6\mu Wb=6\times 10^{-6}weber$

We know that $\Phi =BACOS\Theta$

So $6\times 10^{-6}=3\times 10^{-3}\times 27.712\times 10^{-4}cos\Theta$

$cos\Theta =0.7217$

$\Theta =43.80^{\circ}$

You might be interested in

A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the

Shkiper50 [21]

Answer:

ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round = 144 Kg

angle = 33.2°

moment of inertia

$I = \dfrac{1}{2} m R^2$

$I = \dfrac{1}{2}\times 144 \times 2.75^2$

I = 544.5 kg.m²

torque = force x radius

τ = 37.7 x 2.75

τ = 103.675 N.m

angular acceleration

$\alpha= \dfrac{\tau}{I}$

$\alpha= \dfrac{103.675}{544.5}$

α = 0.190 rad/s²

now ,

distance = $33.2\times \dfrca{2\pi}{360}$

d = 0.579 rad

we know,

using equation of rotational motion

$d = \omega t + \dfrac{1}{2}\alpha t^2$

$0.579 = \dfrac{1}{2}\times 0.190\times t^2$

t = 2.46 s

angular speed

ω = α x t

ω = 0.19 x 2.46

ω = 0.467 rad/s

7 0

2 years ago

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

aleksandrvk [35]

Hope this helps!

-Lilly

3 0

3 years ago

When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the

Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

Voltage drop across the wire V = 0.55 volt

We know that resistance is given by $R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega$

Length of the copper wire l = 1 m

Resisitivity of the copper wire $\rho =1.72\times 10^{-8}\Omega m$

We know that resistance $R=\frac{\rho l}{A}$

$0.00177=\frac{1.72\times 10^{-8}\times 1}{A}$

$A=969.4545\times 10^{-8}m^2$

As area $A=\pi r^2$

$3.14\times r^2=969.4545\times 10^{-8}$

$r=17.57\times 10^{-4}m$

So diameter $d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m$ = 351.42 mm

3 0

3 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0

3 years ago

A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.28). A light beam of variable wavelength from air i

Makovka662 [10]

Answer:

Explanation:

This is case of interference in thin films

for constructive interference in thin film the condition is

2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .

2 x 1.28 x t = λ/2 , if n = 0

2 x 1.28 x t = 605 /2

t = 118.16 nm .

the minimum non-zero thickness of the oil film required = 118.16 nm.

8 0

3 years ago