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2 years ago

How does convection current helps cooling the system of engines

1 answer:

6 0

As the engine heats up, a natural circulation starts, as coolant rises through the engine block by convection. It passes through the top hose, and into the radiator. Inside the radiator, heat is removed from the coolant as it falls from the top to the bottom.
Hope this helps!

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A student shakes a rope such that 20 complete vibrations are made in 4.00 seconds. Determine the vibrational frequency of the ro

fgiga [73]

Answer:

The vibrational frequency of the rope is 5 Hz.

Explanation:

Given;

number of complete oscillation of the rope, n = 20

time taken to make the oscillations, t = 4.00 s

The vibrational frequency of the rope is calculated as follows;

$Frequency = \frac{number \ of \ complete \ vibrations}{time \ taken} \\\\Frequency = \frac{20 }{4 \ s} \\\\Frequency = 5 \ Hz$

Therefore, the vibrational frequency of the rope is 5 Hz.

8 0

2 years ago

Driving Zone 7 corresponds with

stepan [7]

Your driving zone refers to the areas of space around your car, it refers to all the area around your car as far as your eyes can see.
Each car has seven zones numbered from 1 to 7. Driving zone 7 corresponds with THE SPACE YOUR VEHICLE IS OCCUPYING. The other zones are as follows:
zone 1 = area directly infront of your car
zone 2 = your left lane
zone 3 = your right lane
zone 4 = left rear of your car
zone 5 = right rear of your car
zone 6 = area directly behind your car.
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3 0

2 years ago

The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find

snow_tiger [21]

Answer:

The emf of the generator is 6V

The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7 + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7 + 0.3r

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

6 0

2 years ago

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

1 year ago

A 3 column table with 5 rows labeled table A. The first column is labeled voltage in volts with entries 1.0, 5.0, 10, 20, 50. Th

Volgvan

Answer:

0.25A

1.0A

2.5A

Explanation:

this is only for the calculated current column on edge

6 0

3 years ago

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