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3 years ago

What is an example of vaporization?

Physics

2 answers:

Alenkasestr [34]3 years ago

7 0

Answer:

just search it up you'll get ur answer

Phoenix [80]3 years ago

5 0

Boiling water is what I would put

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A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end tex

love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.

For this "type" of motion, displacement (Δx) can be determined by:

$\Delta x = v_{i}.t + \frac{a}{2}.t^{2}$

$v_{i}$ is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

$\Delta x = 5(6) - \frac{4}{2}.6^{2}$

$\Delta x = 30 - 2.36$

$\Delta x$ = - 42

Displacement of the boat for t=6.0s interval is $\Delta x$ = - 42m, i.e., 42 m to the left.

8 0

4 years ago

The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is

almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

$I = I_{0}e^{\frac{t}{\tau\\ } }$

$\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }$

Apply inverse logarithmic function:

$ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}$

The final form will be:

$\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}$

Inputing values for I, IO, and t:

$\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95$

3 0

4 years ago

One ounce of a well-known breakfast cereal contains 104 Calories (1 food Calorie = 4186 J). If 2.5% of this energy could be conv

GalinKa [24]

Answer:

482.4 kg

Explanation:

104 calorie = 435344 J

The 2.5% of this energy equals to

435344 * 0.0025 = 10883.6 J

If this energy is converted to work done on lifting a barbell by a distance of 2.3. By the formula of Work W = force * distance then the force he exerts on the barbell is

$F = \frac{W}{D} = \frac{10883.6
}{2.3} = 4732 N$