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Maru [420]
2 years ago
5

What is a convex lens?​

Physics
2 answers:
Leona [35]2 years ago
6 0

<em>The convex lens is a lens that converges rays of light that convey parallel to its principal axis (i.e. converges the incident rays towards the principal axis) which is relatively thick across the middle and thin at the lower and upper edges. The edges are curved outward rather than inward.</em>

Lana71 [14]2 years ago
6 0
A lens bound by two spherical surfaces bulging outwards and a single piece of glass that curves outward and converges the light incident on it are both convex lenses
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A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta
alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = I_t

Rotational inertia (I_r) of the record = 0.61\times I_t

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as  \omega _i\  ,\  \omega_f respectively.

Note :

Angular momentum (L) = Product of moment of inertia  (I)  and angular velocity (\omega) .  

Lets say,

⇒ initial angular momentum = final angular momentum

⇒  L_i=L_f

⇒ (I_t)\times \omega_i = (I_t+I_r)\times \omega_f

⇒ \omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i) ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ K_i=\frac{I_t\times \omega_i^2}{2}       ⇒ K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}

Their ratios:

⇒ \frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }    

⇒ \frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}

Plugging the values of  \omega _f^2 as \omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2 from equation (i) in the ratios of the Kinetic energies.

⇒ \frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}

Now,

The Kinetic energy lost in fraction can be written as:

⇒ \frac{K_f-K_i}{K_i}

Now re-arranging the terms.

\frac{K_f-K_i}{K_i}  =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}

Plugging the values of  I_r and I_t .

⇒ \frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788

To find the percentage we have to multiply it with 100 and here negative means for loss of Kinetic energy.

⇒ \frac{K_f}{K_i} = =-0.3788\times 100= 37.88

So the percentage of the initial Kinetic energy lost is 37.88

4 0
2 years ago
A substance that accelerates the rate of a chemical reaction is called a(an)
Dahasolnce [82]
<span>A substance that accelerates the rate of chemical reaction is called a catalyst. It serves as an alternative pathway for the reaction product. The increase of rate of chemical reaction is because catalyst has low activiation energy than the original pathway. The low activation energy will increase the amount of molecules that can benefit in the energy created by the catalyst.</span>
5 0
3 years ago
Help please! this is physics !
KATRIN_1 [288]

Answer:

4. The choose b. 0.000355

Ans; 3.55× 10-⁴ = 0.000355

5. The choose C. 80600

Ans; 8.06 ×10⁴= 806×10² = 80600

I hope I helped you^_^

4 0
2 years ago
Read 2 more answers
In a convex lense f=20cm, m=1,then what is u and v?​
katen-ka-za [31]

Answer:

I'm not sure if I know whatever the answer is

3 0
2 years ago
Please help ASAP.
Bess [88]

Im not 100% sure you have to tell me if im wrong or not.

D

B

C

3 0
3 years ago
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