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White raven [17]
1 year ago
6

2) What did Rutherford discover in his experiment? a) nucleus b) electrons c) neutrons

Physics
1 answer:
OlgaM077 [116]1 year ago
5 0
A) nucleus because some aims reflected off the gold foil proving that there was something bigger inside the atom hence discovering the nucleus
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A battery is replaced with one of lower emf. State and explain how the resistance of the lamps would have to change in order to
kow [346]

The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.

The current flowing through the lamp is given by Ohm's law:

I=\frac{V}{R}

where

V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.

The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.

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3 years ago
Water is moving at a velocity of 2.00 m/s through a hose with an Internal diameter of 1.60 cm. What is the volume flow rate at t
Leya [2.2K]

Answer:

15 m/s

Explanation:

3 0
2 years ago
The particle in the atom with a negative charge is the ______<br> Answer here
NISA [10]

Answer:

Explanation:

The electron has a negative charge. Proton is positive and neutron is neutral.

4 0
3 years ago
Read 2 more answers
The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part
svetoff [14.1K]

Answer:

B. distance/potential

Explanation:

Quizlet

5 0
3 years ago
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0
3 years ago
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