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3 years ago

A rectangular loop consists of 80 turns, each carrying a current 8

Physics

1 answer:

liberstina [14]3 years ago

3 0

Magnetic moment of loop is given by the formula

$M = NiA$

$M = 80*8 *0.4*0.3$

$M = 76.8$

now magnetic field is given here

$B = 40 * 10^{-3} T$

Now the torque on the loop is given by

$T = MBsin\theta$

$T = 76.8* 40*10^{-3}*sin30$

$T = 1.536 Nm$

so above is the torque on the loop.

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Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe

Ede4ka [16]

The final velocity ( $v_1_f$ ) of the first astronaut will be greater than the final velocity of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

The given parameters;

Mass of the first astronaut, = m₁
Mass of the second astronaut, = m₂
Initial velocity of the first astronaut, = v₁
Initial velocity of the second astronaut, = v₂ > v₁
Mass of the ball, = m
Speed of the ball, = u
Final velocity of the first astronaut, = $v_f_1$
Final velocity of the second astronaut, = $v_f_2$

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

$m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f$

if v₂ > v₁, then $v_1_f > v_2_f$ , to conserve the linear momentum.

Thus, the final velocity ( $v_1_f$ ) of the first astronaut will be greater than the final velocity of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

5 0

3 years ago

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each

irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

$\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s$

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

$\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s$

So the flow speed of each of the narrower pipe is:

$v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}$

$v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s$

8 0

3 years ago

During science class, while studying mixtures, you mix together iron filings and sand. Your teacher challenges you to separate t

Lapatulllka [165]

Answer : Use a magnet to pull out the iron filings as they are attracted to a magnet

Explanation : we use the magnet to separate the sand from the iron filing. Because the magnet has an attraction power.

We can say that we can use the magnet to attract the iron filings out of the mixture because iron is magnetic solid, but sand will not attract because sand is not magnetic solid.

So, we use a magnet to pull out the iron filings as they are attracted to a magnet.

8 0

3 years ago

Read 2 more answers

A 50.0 kg child stands at the rim of a merry-go-round of radius 1.50 m, rotating with an angular speed of 3.00 rad/s. (a) what i

White raven [17]

Weight of the child m = 50 kg
Radius of the merry -go-around r = 1.50 m
Angular speed w = 3.00 rad/s
(a)Child's centripetal acceleration will be a = w^2 x r = 3^2 x 1.50 => a = 9 x
1.5
Centripetal Acceleration a = 13.5m/sec^2
(b)The minimum force between her feet and the floor in circular path
Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424
Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09
Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s
Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2

/ 1.5
F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N
(c)Minimum coefficient of static friction u
F = u x m x g => u = F / m x g => u = 675/ 50 x 9.81 => 1.376
u = 1.376
Hence with the force and the friction coefficient she is likely to stay on merry-go-around.

8 0

3 years ago

Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t

Alex_Xolod [135]

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

$B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}$

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

$B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\ T$

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

5 0

3 years ago