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Svet_ta [14]
3 years ago
8

Different between current and electrons?

Physics
1 answer:
BlackZzzverrR [31]3 years ago
3 0

Answer:

Simply,

<u>electrons</u> are "PARTICLES" orbiting the atoms, where, <u>current</u><u> </u>is the FLOW of some (free-to-move-around) electrons in a wire...

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The diagram shows an example of convection.
const2013 [10]

Answer: Option (C) is the correct answer.

Explanation:

When we heat a fluid then the movement within the fluid makes hot (less dense) material to rise and cooler (more denser) material to sink at the bottom. This process is known as convection.

Thus, in the diagram hot (less dense) water will rise and cooler (more dense) water sinks at the bottom.

Therefore, we can conclude that according to the arrow the label belongs to cooler water sinks.

5 0
3 years ago
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Two resistors, R1 and R2, are
Aleonysh [2.5K]

Answer:

The value of R₂ is equal to 24.75 ohms.

Explanation:

Given that,

Two resistors, R₁ and R₂, are  connected in parallel.

The equivalent resistance is 14.5 ohms

We need to find the value of R₂.

When two resistors are connected in parallel. The equivalent resistance is given by :

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{14.5}=\dfrac{1}{35}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_2}=\dfrac{1}{14.5}-\dfrac{1}{35}\\\\\dfrac{1}{R_2}=0.04039\\\\R_2=\dfrac{1}{0.04039}\\\\R_2=24.75\ \Omega

So, the value of R₂ is equal to 24.75 ohms.

8 0
3 years ago
A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength eq
Veronika [31]
F = kq1q2/r<span>2
Where,
F - Coulomb Force
k - constant value which is equal to </span>8.98 × 10^9<span> newton square metre per square coulomb
q1 and q2 -  two electric charges
r - distance.

5.8 * 10^5 = 1.5 * 10^-9 / r^2
</span><span>5.8 * 10^5 r^2 = 1.5*10^-9
</span>r^2 = 0.0000258620
r = 0.0050854694

So the distance is equal to 5.09 x 10^-3 
3 0
3 years ago
Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Vedmedyk [2.9K]

Answer:

(a) V (A) =  0.7 m/s,

(b) V (A) =  0.7 m/s,

(c) V (B) =  0.7 m/s

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a)  law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V      (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V =  0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) =  0.7 m/s

b)  V(A) =  0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) =  0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s =  3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1)  ( dividing both side by 3.50 Kg)

For elastic collision  

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u  ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

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3 years ago
Which changes will increase the rate of reaction during combustion? decreasing the area of contact between the reactants adding
STALIN [3.7K]
The correct answer is B. The addition of oxygen will increase the rate of combustion. oxygen and raw material (fuel) are the substrates for the combustion process, therefore, increase in the same will obviously increase combustion rate. Carbon in the fuel is converted into carbon dioxide and heat. 
3 0
2 years ago
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