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4 years ago

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A round loop of wire carries a current of 100 A, has a radius of 10 cm, and its normal (vector) makes an angle of 30∘ with a mag

netic field of 0.524 T. 1) What is the torque on the loop? (Express your answer to two significant figures.)

1 answer:

Oliga [24]4 years ago

7 0

Answer:

0.823 Nm

Explanation:

current, i = 100 A

radius, r = 10 cm

Angle between the normal and the magnetic field, θ = 30°

Magnetic field, B = 0.524 T

Torque is defined as the

$\tau = i \times A\ B\times sin\theta$

$\tau = 100 \times 3.14\times 0.1\times 0.1\ 0.524\times 0.5$

Torque = 0.823 Nm

Thus, the torque is 0.823 Nm.

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What are some possible examples of genetic characteristics that can be passed down to child??????

elena-s [515]

They can have a close similar appearance to the parents, have close relation of child reactions.

for example, everyone born in my father's side of the family had the tendency to bump their head on something as they fall asleep up to the point when you are a toddler.

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3 years ago

Read 2 more answers

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago

A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the

Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁ = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

m₁v₁ + m₂v₂ = v(m₁ + m₂)

Insert parameters and solve for v;

5 x 1 + 15 x 0 = v (5 + 15 )

5 = 20v

v = $\frac{5}{20}$ = 0.25m/s

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3 years ago

An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa

Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'

Now,

the Doppler effect is given as:

$f'=f\frac{v\pm v_o}{v\pm v_x}$

where, v is the speed of sound

since the ambulance is moving towards the cyclist. thus, the sign will be positive

thus,

$v_x=\frac{f}{f'}(v+v_o)-v$

on substituting the values, we get

$v_x=\frac{1790}{1780}(343+2.36)-343$

or

vₓ = 4.30 m/s

Hence, <u>the speed of the ambulance is 4.30 m/s</u>

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3 years ago

You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of yo

luda_lava [24]

Answer:

75.84%

Explanation:

We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.

v=80mph= 35.76 m/s

The radius of curvature is given as , r = 540 m

✓ the normal weight can be denoted as Wn

✓ the apparent weight of the person can be denoted as Wa

Wn= normal weight= mg

Wa=apparent weight = (mg - mv^2/r)

g= acceleration due to gravity= 9.8m/s^2

The apparent weightand normal weight has a ratio of

Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)

If we simplify eqn(1) we have

Mn/Ma=[g - vr^2/g].............eqn(2)

Then substitute the given values

Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8

=0.758×100%

Mn/Ma=75.84%

Hence, the required fraction is 75.84%

4 0

3 years ago