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Bogdan [553]
3 years ago
5

Is this diagram important in exams?

Physics
1 answer:
pickupchik [31]3 years ago
6 0
Yes it might come as a blank diagram and ask you to fill in the blanks or it might ask you to say how much energy is wasted
                                       good luck on your exam
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What's is the kinetic energy of a .235-kg baseball thrown at 50.0m/s
san4es73 [151]

Answer:

34.51

Explanation:

k=1/2mv² is the kenetic energy equation to fill is in

k=[1/2(0.235)×50]²

6 0
2 years ago
When adjusting the valve clearances on the engine, Technician A says the valves are closed when the valve lifter or follower is
andre [41]

Answer:

A. Technician A only

Explanation:

The camshaft opening ramp is what opens the valve, it starts at the basement of the inner circle and closes at the nose while the camshaft closing ramp allows for closing a valve starts at the nose and closes at the inner base circle. Valve clearance on an engine are performed when closing a valve, and is completed when the valve lifter or follower is on the inner base circle of the camshaft lobe.

7 0
3 years ago
PLZ HELP!!!!
zavuch27 [327]

Answer:

AAAAAA!!!!!!!

Explanation:

Hope that helped lol!

8 0
2 years ago
Read 2 more answers
Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q /  (5 x 10⁻² )²

E₁  = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁  = KQ x 10⁴  i / 25  

Similarly electric field due to charge 3Q near 2Q

=  3KQ x 10⁴  i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴  j / 25

Similarly electric field due to charge 4Q near 2Q

=  4KQ x 10⁴  j / (25 x 2 )

= 2 KQ x 10⁴  / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴  ( i + j )  / 25 x √2

Total field =  KQ x 10⁴  i / 25 + 3KQ x 10⁴  j / 25 + 2 KQ x 10⁴  ( i + j )  / ( 25 x √2 )

= KQ x 10⁴  [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j )  N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j )  x 2Q  N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0
3 years ago
Compare and contrast CLF and LED bulbs.
Arte-miy333 [17]

Explanation:

Comparison Chart

Basis For Comparison CFL Bulb LED Bulb

Definition The CFL is defined as the lamp which uses the low pressure mercury vapour gas for producing the visible light It is a PN junction diode which produces light when current passes through it in the forward direction.

Stand For Compact Fluorescent Lamp Light Emitting Diode

Working principle The electrons of mercury when excited emits ultraviolet light. This ultraviolet light when strikes with the fluorescent coating glass converts it into the visible light. The semiconductor material which emits light with the passage of current.

Power Consumption More Less

Mercury Contain Not contain

Destruction Difficult Easy

Efficient Less More

Ballast Requires Not requires

Lifespan 10,000 hours 50,000 hours and more.

Turn on Intensity Delay Quick

Expensive Less More

4 0
2 years ago
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