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2 years ago

What is 28 degrees Celsius converted to Fahrenheit?

Physics

2 answers:

suter [353]2 years ago

6 0

I am great at this the answer is 82.4

torisob [31]2 years ago

3 0

28 degrees celsius converts to 82.4 degrees in Fahrenheit :)

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Part AIf the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negativ

Nana76 [90]

Answer:

Part a: The potential at point 3 and point 6 are V and 0 respectively.

Part b: The charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c: The net charge is 6CV.

Part d: The equivalent capacitance is 6C

Explanation:

As the question is not given here ,the complete question is found online and is attached herewith.

Part a:

V1 = V, V3 = V1 = V

and, V6 = V1-V = 0

The potential at point 3 and point 6 are V and 0 respectively

Part b

Q1 =CV = Q,

Q2 = 2C *V = 2Q

Q3 = 3 C*V = 3Q

So the charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c

Total charge of the system,

Q_net =Q1+Q2+Q3= (1+2+3) CV = 6 CV

So the net charge is 6CV.

Part d:

Equivalent Capacitance = Net charge / Voltage

Eq. C = 6CV/V = 6C

So the equivalent capacitance is 6C

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3 years ago

Solar-powered cars use energy from the Sun to work. A panel on the car absorbs light energy from the Sun, which then generates a

Over [174]

Solar energy, electrical energy, and mechanical energy.

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3 years ago

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2-Pema runs

Oxana [17]

Answer:

20 m/s

Explanation:

Distance covered by man =500m

Time taken by man=25 seconds

Speed of man = distance/time

= 500/25

=20 m/s

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2 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with

Tanya [424]

Answer:

(a) $K_{small}=4839.3J$

(b) $K_{larger}=17406.4J$

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

$K=\frac{1}{2}Iw^2$

Where I is rotational of inertia of solid cylinder about its central axis.

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2$

Substitute the given values

$K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J$

For Part (b)

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2$

Substitute the given values

$K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J$

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3 years ago

Which of the following best describes a plane?

katrin [286]

B, the surface of a flat table.

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3 years ago

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