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Galina-37 [17]
3 years ago
5

In our lab experiment on Ohm's law, the power supply is connected to a circuit containing one resistor, and a direct current of

1.5 A is drawn. If you connect a second identical resistor to the power supply in parallel with the first, (similar to what we did in the lab) how much current is drawn from the battery?
(A) 3.0 A
(B) 0.75 A
(C) 1.5 A
(D) 6.0 A
Physics
1 answer:
lapo4ka [179]3 years ago
3 0

Answer: (A) 3.0=A

Explanation: In order to explain this problem we have to use the OHM law, given by: V=R*I

Besides, we have to consider the resitance equivalent for a parallel connection. This is given by:

1/Re=1/R1+1/R2

If we connect the same resistance, the equivalent resistance is R/2.

Initlally  the current is 1.5 A when one resistance is connected to the batttery. When a second resistance with the same value is connected in parallel to the battery, we have V=Re*Ifinal= (R/2)*Ifinal

also we know that V=R*Iinitial so Iinitial=V/R

then Ifinal= 2*V/R=2*Iinitial

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3 0
3 years ago
True or false<br> Hydrogen fuel cells generate electricity by combining hydrogen with oxygen.
adoni [48]

Answer:

The answer is: True.

Explanation:

If free electrons or other substances could travel through the electrolyte, they would disrupt the chemical reaction. Whether they combine at anode or cathode, together hydrogen and oxygen form water, which drains from the cell. As long as a fuel cell is supplied with hydrogen and oxygen, it will generate electricity.

(Credit: Google)

7 0
3 years ago
A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m
lbvjy [14]

The equation 15v_{i} + 2*0 = (15 + 2)v_{f} (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.  

The horizontal momentum is given by:

p_{i} = p_{f}

m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}

Where:

  • m₁: is the mass of the lab cart = 15 kg
  • m₂: is the <em>mass </em>of the object dropped = 2 kg
  • v_{1}_{i}: is the initial velocity of the<em> lab cart </em>
  • v_{2}_{i}: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
  • v_{1}_{f}: is the final velocity of the<em> lab cart </em>
  • v_{2}_{f}: is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

15v_{1}_{i} + 2*0 = v_{f}(15 + 2)

Therefore, the equation 15v_{i} + 2*0 = (15 + 2)v_{f} represents the horizontal momentum (option 3).

Learn more about linear momentum here:

  • brainly.com/question/2141713?referrer=searchResults
  • brainly.com/question/2400186?referrer=searchResults

I hope it helps you!            

4 0
3 years ago
A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T
den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

m_pv_p=v_c(m_p+m_b) where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

0.5(m_p+m_b)v_c^{2}=0.5kx^{2} where k is spring constant and x is the compression of spring. Substituting the given values then

(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m

7 0
3 years ago
An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k
zheka24 [161]

Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.

Explanation:

1) Assuming that no external forces act during the collision, total momentum must be conserved. As initially the total mass was at rest, so initial momentum is zero, final momentum of all the system must be 0 also.

2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:

m₁ + m₂ + m₃ = M = 14.6 Kg = 4.9 Kg + 5.5 Kg + m₃

Solving for m₃, we have:

m₃ = 14.6 Kg - 4.9 Kg -5.5 Kg = 4.2 Kg.

3) and 4)

As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.

So, we can write two equations, one for the x-component, and other for the y-component, as follows:

pₓ = m₁. v₁ₓ + m₂.v₂ₓ + m₃.v₃ₓ = 0

py = m₁.v₁y + m₂. v₂y + m₃. v₃y =0

Replacing by the values, and solving for v₃ₓ and v₃y, we get:

v₃ₓ = 15.4 m/s

v₃y = 12.9 m/s

v = √(15.4)²+(12.9)² = 20.1 m/s

5) As the center of mass must move as if all the mass were concentrated in this point, and we know that the total momentum must be 0, this tells us that the magnitude of the velocity of the center of mass must be 0 too.

6) As initial kinetic energy is 0, as  the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)

Replacing by the values, we get:

K= 3.62 KJ

4 0
3 years ago
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