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2 years ago

I need help with this problem :(

Physics

1 answer:

Reil [10]2 years ago

7 0

Answer:

it tells you that the speed increases until about 20 seconds then keeps a steady pace for 20 seconds then the speed drops and stops at 55 seconds in the process.

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A university is planning to send an optical telescope into orbit. stuents attending the university will be able to view which of

kodGreya [7K]

Where are the answers? If there's anything about a white light coming from space I would choose that one.

3 0

3 years ago

Read 2 more answers

A 66-foot-tall woman walks at 55 ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th

I am Lyosha [343]

Answer:

a. $\frac{dx}{dt}=20ft/s$

b. $\frac{d(x+L)}{dt}==25ft/s$

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

$tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}$

Solve to find the rate relation

$x=\frac{24}{6}*L$

$x=4*L$

Now the rate of the change rate

$\frac{dx}{dt}=4*\frac{dL}{dt}$

$\frac{dx}{dt}=4*5ft/s=20ft/s$

Finally the rate of her shadow moving

$\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}$

$\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s$

5 0

3 years ago

Newtons laws of motion

Viktor [21]

Answer:

Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.

Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.

Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.

7 0

3 years ago

An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe

nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

$f'=\frac{(v+v_o)}{(v+v_s)}f$

Where f is the actual frequency, $f'$ is the observed frequency, $v$ is the velocity of the sound waves, $v_o$ the velocity of the observer (which is negative if the observer is moving away from the source) and $v_s$ the velocity of the source (which is negative if is moving towards the observer). For this problem: