Answer is C. Kelvin, Kelvin is the temperature scales does water boil at the highest numerical value. Hope it helped you, and have a great day.

-Charlie

7 0

3 years ago

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A 50.0-kg crate is being pulled along a horizontal smooth surface. The pulling force is 10.0 n and is directed 20.0° above the h

steposvetlana [31]

Answer:

$0.188 m/s^2$

Explanation:

Assuming the crate does not lift above the ground and remains along the floor, then its acceleration will be in the horizontal direction. Therefore, we can use Newton's second law to find its acceleration:

$F_x = ma_x$

where

$F_x$ is the net force on the crate along the x-direction

m is the mass of the crate

$a_x$ is the acceleration

Here we have:

m = 50.0 kg

$F_x = F cos \theta = (10.0 N)(cos 20.0^{\circ})=9.4 N$ is the component of the pulling force along the horizontal direction

Solving for the acceleration,

$a_x = \frac{F_x}{m}=\frac{9.4}{50.0}=0.188 m/s^2$

6 0

3 years ago

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

babunello [35]

Answer:

Separation increases at all times that rock X falls because it falls with a greater speed

Explanation:

For both rocks, let initial velocity ∪=0

To find the displacement at any given time interval of Δt then

S= ∪Δt +0.5gΔt²

Since rock X is first released followed by Y, then X has a greater speed than Y therefore the distance covered by X is longer. This is because despite 0.5gΔt² being same for both rocks at any time Δt but rock X having already attained some velocity, its ∪Δt is more hence the separation S increases. Conclusively, S increases at all times that rock X falls since rock X falls with a greater velocity than rock Y

7 0

3 years ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

3 years ago