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kupik [55]
3 years ago
9

What is the focal length of the mirror? follow the sign conventions?

Physics
1 answer:
Ipatiy [6.2K]3 years ago
4 0
Well, the focal length is a distance to the point in which all of the rays of a parallel beam cross. Since a parallel beam reflected by the mirror is still parallel and does not converge, no matter how far from the mirror it travels, the focal distance of a mirror is

f = -∞

minus sign denotes that the image in the mirror is imaginary
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When the chemical name of a compound is being written, the subscripts will determine the symbols. elements. prefixes. suffixes.W
Xelga [282]

Answer:

C. Prefixes

Explanation:

I took the quiz on edge

5 0
3 years ago
Read 2 more answers
A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How
alina1380 [7]

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2

(a)

The final angular speed is given as;

\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 +  2\alpha \theta\\\\\omega _f^2 =   2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f  = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s

(b) the time taken to turn 5.5 revolutions is given as

\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s

3 0
2 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth’s center. Satellite A is to o
Vera_Pavlovna [14]

Answer:

Explanation:

Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km

Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km

Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite

Orbital potential energy of a satellite A = - GMm / Ra

Orbital potential energy of a satellite B = - GMm / Rb

PE of satellite B /PE of satellite A

=  Ra / Rb

= 12740 / 25480

= 1 / 2

b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same

KE of satellite B /KE of satellite A

= 1 / 2

c ) Total energy will be as follows

Total energy = - PE + KE

- P E + PE/2

= - PE /2

Total energy of satellite B / Total energy of A

= 1 / 2

Satellite B will have greater total energy because its negative value is less.

5 0
3 years ago
3. Rock A is thrown horizontally off of a cliff with a velocity of 20 m/s. The
Anna11 [10]

Answer:

44.1 m

Explanation:

4 0
2 years ago
An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo
IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0
3 years ago
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