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3 years ago

15

Two identical objects are moving directly toward one another at the same speed v. ~v â~v m m what is the total kinetic energy of

the system of the two objects?

1 answer:

Tresset [83]3 years ago

3 0

If they are identical, then their masses are the same. The speeds are the same. KE has no direction.
Total KE = 1/2 m v^2 + 1/2 mv^2 = mv^2

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Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

QveST [7]

Answer:

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Explanation:

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2 years ago

The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a

Degger [83]

Answer:

147.456077993 Hz

Explanation:

$f_0$ = Frequency of the sonar = 22 kHz

$v_w$ = Velocity of the whale = 4.95 m/s

v = Speed of sound in water = 1482 m/s

The difference in frequency is given by

$\Delta f=f_0\times\dfrac{2v_{w}}{v-v_w}\\\Rightarrow \Delta f=22000\times\dfrac{2\times 4.95}{1482-4.95}\\\Rightarrow \Delta f=147.456077993\ Hz$

The difference in frequency is 147.456077993 Hz

6 0

3 years ago

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

2 years ago

the minimum flying speee for a bird called a house martin is 9.0ms^-1. it reaches this speed by falling from its nest before swo

Lilit [14]

Answer:

The nest must be about 4.15 meters above ground

Explanation:

Use the velocity equation under accelerated motion (acceleration of gravity ):

$v_f=v_i+a\,*\,t$

which for this case has initial velocity = 0 (falls from the nest), final velocity = 9 m/s, and a = 9.8 m/s^2, then we can find the time needed in air while falling to reach the required speed:

$v_f=v_i+a\,*\,t\\9=0+9.8\,t\\t=\frac{9}{9.8} \, sec\\t \approx 0.92\,\, sec$

We now use this time value to find the distance covered in free fall during 0.92 seconds:

$d=\frac{1}{2} \,9.8 \,t^2=4.9\,(0.92)^2=4.147 \,meters\approx 4.15\,meters$

8 0

2 years ago

a cyclist applies a constant forward force of 20 N to maintain a velocity 2.5 m/s. how much power does the cyclist deliver?

forsale [732]

The power exerted by the cyclist is determined as 50 W.

<h3>Average power exerted by the cyclist</h3>

The power exerted by the cyclist is calculated as follows;

P = FV

where;

F is the applied force
V is velocity

P = 20 x 2.5

P = 50 W

Thus, the power exerted by the cyclist is determined as 50 W.

Learn more about power here: brainly.com/question/25263760

#SPJ1

4 0

2 years ago