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Irina-Kira [14]
2 years ago
9

If someone were to drop an orb from a height of 34,000 meters approximately, and then drop a second orb at 34,100 meters, estima

te how long it would take for the second orb to reach the ground if it takes the first orb 55~ seconds to impact.
Extra Info:
-The first orb weights 2 pounds and the second orb weights only 2.5 pounds.
-The ground is entirely flat and level. (as if you smashed a flattening board into the ground)
-They are both dropped with no force put into it, simply held up and let go of.
-Both Orbs are the same size, just different density.
-The heights are estimates and are not exact, hence the answer can be an estimated one.

(this question sparked from boredom ._.)
If info required I can provide.
Physics
1 answer:
Pavlova-9 [17]2 years ago
5 0

Answer:

Explanation:

SO the ball falls 618 meters per second meaning that's how fast 2 pounds fall per second so i have to find out how fast .5 pounds falls which i think its 154 meters per second per .5 pounds meaning the second orb would fall 772 meters per second so the second orb would take 44.1 seconds to hit on impact. Im not sure if this is right but this is my best guess.

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If an object is projected upward with an initial velocity of 127 ft per? sec, its height h after t seconds is h equals negative
Stolb23 [73]
To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:

h = -16t^2 + 127t

We substitute 55 seconds to t and obtain,

h = -16(55)^2 + 127(55)
h = - 41415
4 0
3 years ago
PLZ HELP NOW
ZanzabumX [31]

Answer: mass x height x gravitational field strength (g)

note: gravitational field strength (g) = 10 N/Kg

55 x 15 x 10 = 8250

gpe = 8250j

Explanation:

4 0
3 years ago
The Earth orbits the Sun at a speed of 30 km/s. At that speed it completes one path around the Sun every year. Of course, as tha
romanna [79]

Answer:

a. 299,792,458 m/s

Explanation:

Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.

3 0
3 years ago
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

3 0
3 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
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