(b) Always act on the different bodies in opposite directions
complete question:
An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot
Answer:
a ≈ 5281 ft
Explanation:
The observer at the top of a 462 ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°.
The angle of depression form the top of the cliff = 5°
The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°. 90° - 5° = 85° Note sum of an angle on a right angle is 90°.
using SOHCAHTOA principle we can solve for the distance from the base of the cliff to the point on the ground(a)
tan 85° = opposite / adjacent
tan 85° = a / 462
cross multiply
462 × tan 85° = a
a = 11.4300523 × 462
a = 5280.66 ft
a ≈ 5281 ft
- The potential difference between two locations in an electric circuit is measured using a voltmeter.
- If the electricity passes through the voltmeter it shows deflection.
<h3>What is the purpose of a voltmeter?</h3>
- A voltage meter, usually referred to as a voltmeter, is a device that measures the voltage, or potential difference, between two points in an electrical or electronic circuit.
- volts is the unit of voltmeter(volts, millivolts, kilovolts)
<h3>What is the explanation for the link between current and voltage?</h3>
- Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.
<h3>What is ohm's law in circuit?</h3>
- V = IR, where V is voltage, I is current, and R is resistance, is known as Ohm's Law.
- If you know the voltage of the battery in the circuit and how much resistance is in the circuit, you may use Ohm's Law to identify properties of a circuit, such as how much current is flowing through it.
To learn more about current and voltage visit:
brainly.com/question/10254698
#SPJ4
Explanation :
Absorption coefficient of a material determines how much sound is absorbed by the material.
To build a soundproof room, Heavy curtains and carpet can be used. They reduce reverberation.
Reverberation means an echoing sound which persists for some time. For example, when we bang on a huge piece of metal, we hear the reverberation even after we stop banging.
Hence, option (A) and (D) are correct.
This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
Height at any time 'T' after the drop =
(initial height) +
(initial velocity) x (T) +
(1/2) x (acceleration) x (T²) .
For the balloon problem ...
-- We have both directions involved here, so we have to define them:
Upward = the positive direction
Initial height = +150 m
Initial velocity = + 3 m/s
Downward = the negative direction
Acceleration (of gravity) = -9.8 m/s²
Height when the bag hits the ground = 0 .
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)
-4.9 T² + 3T + 150 = 0
Use the quadratic equation:
T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]
= (-1/9.8) [ -3 plus or minus 54.305 ]
= (-1/9.8) [ 51.305 or -57.305 ]
T = -5.235 seconds or 5.847 seconds .
(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)
Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
H'(t) = v₀ + a T
The extremes of 'H' (height) correspond to points where h'(t) = 0 .
Set v₀ + a T = 0
+3 - 9.8 T = 0
Add 9.8 to each side: 3 = 9.8 T
Divide each side by 9.8 : T = 0.306 second
That's the time after the drop when the bag reaches its max altitude.
Oh gosh ! I could have found that without differentiating.
- The bag is released while moving UP at 3 m/s .
- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
(3 / 9.8) = 0.306 second .
At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .
Distance climbed during that time = (average speed) x (time)
= (1.5 m/s) x (0.306 sec)
= 0.459 meter (hardly any at all)
But it was already up there at 150 m when it was released.
It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.
We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.
I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.
