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galina1969 [7]
3 years ago
8

Plz do all of it i will give brainlest and thanks to best answer plz do it right

Physics
2 answers:
AlladinOne [14]3 years ago
3 0
The answer is a rainforest I’m pretty sure
Alika [10]3 years ago
3 0
Answer :
Rainforest
Explanation:
It should be correct .
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A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet
Rzqust [24]

Answer and Explanation:

This experiment is known as Lenz's tube.

The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:

\varepsilon =-\frac{d\phi_B}{dt}

This emf induced on the surface of the tube generates a current within it according to Ohm's Law:

V=IR

This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.

7 0
3 years ago
Read 2 more answers
What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?
pentagon [3]

Answer:

\tau_a = F a sin \theta

Explanation:

The torque of a force is given by:

\tau = F d sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation of the system

\theta is the angle between the direction of the force and d

In this problem, we have:

F, the force

a, the distance of application of the force from the centre (0,0)

\theta, the angle between the direction of the force and a

Therefore, the torque is

\tau_a = F a sin \theta

5 0
3 years ago
Why does a box on the seat of a car slide around on the c one the car speeds up slows down or turns a corner?
Feliz [49]
Because the box keeps going straight at the same speed, while the seat under it speeds up, slows down, or changes direction.
8 0
3 years ago
A 6.0-v battery maintains the electrical potential difference between two parallel metal plates separated by 1.0 mm. what is the
Gnoma [55]
The voltage<span> difference between the two plates can be expressed in terms of the </span>work<span> done on a positive test charge q when it moves from the positive to the negative plate.</span><span>
E=V/d
where V is the voltage and d is the distance between the plates.
 
So,

E=6.0V/1mm= 6000 V/m. The electric field between the plates is 6000 V/m.</span>
5 0
3 years ago
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0
3 years ago
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