Ask question

Ask question

All categories

3 years ago

13

A wire is formed into a circle having a diameter of 10.3 cm and is placed in a uniform magnetic field of 2.98 mT. The wire carri

es a current of 5.00 A. Find the maximum torque on the wire.

1 answer:

Ksju [112]3 years ago

8 0

Answer:

T(max) = 1.17 × 10⁻⁴Nm

= 117μNm

Explanation:

T = BIA sinθ

A = area enclosed

θ = angle between normal plane

for max. torque θ = 90, (sin90° =1)

T = BIA sin90°

T= BI (πd/4)

T = $T_m_a_x = \frac{1}{4} (2.98 * 10^-^3)(5)(\pi )\\T_m_a_x = 1.17 * 10^-^4Nm\\T_m_a_x = 117UNm$

You might be interested in

A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A

zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

$V_{orbital} = \sqrt{\frac{GM_E}{R}}$

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

$V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}$

$V_{orbital} = 5591.62m/s$

$V_{orbital} = 5.591*10^3m/s$

PART B) The period of satellite is given as,

$T = 2\pi \sqrt{\frac{r^3}{Gm_E}}$

$T = \frac{2\pi r}{V_{orbital}}$

$T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}$

$T = 238.61min$

PART C) The gravitational force on the satellite is given by,

$F = ma$

$F = \frac{1}{4} mg$

$F = \frac{270*9.8}{4}$

$F = 661.5N$

5 0

3 years ago

Why is ice harder than liquid water?

yaroslaw [1]

Hello~

Ice is harder than liquid water because<span> the molecules of ice are linked more tightly together than the molecules of liquid water.

Hope this helps! </span>

6 0

2 years ago

Which factors affect the strength of the electric force between two objects

Margaret [11]

<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>

8 0

3 years ago

Read 2 more answers

Who reported four “element” classifications, but included some substances that were combinations of elements rather than true el

notsponge [240]

Antoine-Laurent Lavoisier was the first person to report the four element classification system but also ended up including some compounds rather than elements.

7 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

2 years ago