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sweet-ann [11.9K]
2 years ago
7

A nonconducting sphere has mass 80.0 g and radius 20.0cm . A flat, compact coil of wire with five turns is wrapped tightly aroun

d it, with each turn concentric with the sphere. The sphere is placed on an inclined plane that slopes downward to the left (Fig. P29.65), making an angle Ф with the horizontal so that the coil is parallel to the inclined plane. A uniform magnetic field of 0.35T vertically upward exists in the region of the sphere.(b) Show that the result does not depend on the value of Ф .
Physics
1 answer:
seraphim [82]2 years ago
8 0

The torque value is 560 Nm, so the result does not depend on the value of Ф.

Given:

Mass=80.0 g

Radius=20.0 cm

Magnetic field=0.35T

In a magnetic field, a loop of current-carrying wire will experience the following torque:

τ=μ×Β

Here, is the μ magnetic moment and Β is the magnetic field.

τ= 1600 × 0.35 T

τ= 560 Nm

<h3>What is torque and its SI unit?</h3>

The rotating equivalent of force is torque. It is also known as the moment, moment of force, rotating force, or turning effect depending on the field of research. It serves as an example of how a force can alter a body's rotational motion.

The Newton-metre, or kgm2sec-2, is the SI unit for torque. How did we get to this point? Considering the equation Torque = Force X Distance Torque is measured in newton-meters, whereas distance is measured in meters and force is measured in newtons.

To know more about torque , visit :

brainly.com/question/561459

#SPJ4

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A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at
docker41 [41]

Answer:

y = 4.36

Explanation:

Let the height of the ladder be L

L = 10

Also:

  • Let x = distance\ from\ the\ base\ of\ the\ ladder
  • Let y = height\ of\ the\ base\ of\ the\ ladder

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

L^2 = x^2 + y^2 --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

x = 9

Substitute 9 for x and 10 for L in L^2 = x^2 + y^2

10^2 = 9^2 + y^2

100 = 81 + y^2

Make y^2 the subject

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y^2 = 19

Make y the subject

y = \sqrt{19

y = 4.36

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0
3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

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