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VMariaS [17]
2 years ago
6

Hi guys! pls help I asked this question almost 2 times and still didn't receive my answers.....Thanks in advance..

Physics
1 answer:
rjkz [21]2 years ago
3 0

Answer:

1. Lateral inversion is a phenomenon in which left appears to be right and vice versa. It is due to direction that light follows when it strikes a reflecting surface, generally a mirror.

These are the letters which don't show lateral inversion A,H,O,T,U

2. USES OF CONCAVE MIRROR

They are used as shaving mirrors to see a larger image of the face.

Dentists use concave mirrors to view large images of the teeth of the patients.

USES OF CONVEX MIRROR

It is is used as a rear view mirror in vehicles.

It is used as a vigilance mirror.

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volume and erosion

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Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
Oksi-84 [34.3K]

-- The net force on the box is 2N to the left.

-- The box will move to the left and accelerate to the left.

-- F=ma . a=F/m . a=(2N)/(4kg).

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2 years ago
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
bazaltina [42]

Answer:

The current in the rods is 171.26 A.

Explanation:

Given that,

Length of rod = 0.85 m

Mass of rod = 0.073 kg

Distance d = 8.2\times10^{-3}\ m

The rods carry the same current in the same direction.

We need to calculate the current

I is the current  through each of the wires then the force per unit length on each of them is

Using formula of force

\dfrac{F}{L}=\dfrac{\mu_{0}I^2}{2\pi d}

\dfrac{mg}{L}=\dfrac{\mu_{0}I^2}{2\pi d}

Where, m = mass of rod

l = length of rod

Put the value into the formula

I^2=\dfrac{mgd}{\mu L}

I^2=\dfrac{0.073\times9.8\times8.2\times10^{-3}}{2\times10^{-7}}

I=\sqrt{29331.4}

I=171.26\ A

Hence, The current in the rods is 171.26 A.

5 0
2 years ago
A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp
AlekseyPX

Answer:

(A) 2.4 N-m

(B) 0.035kgm^2

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque \tau =Fr, here F is force and r is distance

So \tau =20\times 0.12=2.4Nm

(B) Moment of inertia is equal to I=\frac{1}{2}mr^2

So I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2

Torque is equal to \tau =I\alpha

So angular acceleration \alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2

(C) As the disk starts from rest

So initial angular speed \omega _{0}=0rad/sec

Time t = 4.6 sec

From first equation of motion we know that \omega =\omega _0+\alpha t

So \omega =0+68.571\times  4.6=315.426rad/sec

(D) Kinetic energy is equal to KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J

(E) From second equation of motion

\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad

3 0
3 years ago
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