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3 years ago

13

20 points and brainiest

1 answer:

DIA [1.3K]3 years ago

7 0

Answer:

electrical energy sometimes.

Explanation:

125.0m

300 degree Fahrenheit

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Which of the following observations indicates that an object is falling at terminal velocity?

vitfil [10]

Answer: D

Explanation:

When an object falls gravity is pulling down on it and is picking up speed, but as it gains speed air resistance becomes a faster. Air resistance increases with speed. And that force keeps it from accelerating eventually the object will pick up speed such that the force due to air resistance will keep it from getting any more speed at that point force due to air resistance is equal to its weight (mg) and the net force is equal to zero so it won’t accelerate any more at that point it is said to be moving in terminal velocity.

When an object has reached terminal velocity, it will have a constant velocity

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3 years ago

How do the causes of surface and deepwater currents differ

zepelin [54]

The temperature is colder, and the water pressure is higher.

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Which is a freshwater region characterized as a large standing body of water? A. lake B. river C. ocean D. stream

Alexus [3.1K]

This is A.) lake. A river is a small amount of water that isn't always fresh water. A stream is too small. And an ocean is made of salt water.

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3 years ago

Read 2 more answers

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

a car traveling at 30m/s notices an accident about 65m up the road. The car's brakes are capable of decelerating at a rate of 6.

uysha [10]

Yes because you would have at least 3 car spaces

4 0

3 years ago