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3 years ago

If a marble is released from a height of 10 meters how long would it take to hit the ground?

1 answer:

3 0

S = ut + 0.5at^2

10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} 

t^2 ~ 2.04 

t ~ 1.43 seconds

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Say you dropped a cannonball from the 17.0–meter mast of a ship sailing at 2.0 meters/second. How far from the base of the mast

Mice21 [21]

The correct choice is A. 0 meters.

If you simply dropped the cannonball and didn't throw it horizontally,

then it'll fall straight down the mast and land on the deck right next to

the mast.

While you were up there holding it, before you dropped it, the cannonball

was moving horizontally, at 2.0 meters/second, along with the rest of the

ship and everything else aboard. It continued doing that after it dropped,

and from the point of view (in the reference frame) of the mast and everyone

on the ship, it fell straight down, parallel to the mast.

Now that we have that question answered, we can proceed to the more-

important ones. I answered the easy one, but YOU'll have to answer these:

==> WHY did you climb the mast carrying a cannonball ?

Have you been drinking sea water or bad rum ?

==> WHY did you drop it, and never even yell "LOOK OUT BELOW !" ?

==> How many formerly-able-bodied souls were injured by the

plummeting cannonball ?

==> What did everybody ELSE yell after the impact ?

==> What did they do to you after they brought you down ?

3 0

3 years ago

Read 2 more answers

Brainly malfunction?| I'm not sure why but...

Brut [27]

Answer:

its probably still trying to load ur next rank or whatever it did it to me too

Explanation:

6 0

2 years ago

The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is

almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

$I = I_{0}e^{\frac{t}{\tau\\ } }$

$\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }$

Apply inverse logarithmic function:

$ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}$

The final form will be:

$\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}$

Inputing values for I, IO, and t:

$\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95$

3 0

3 years ago

If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that th

olga55 [171]

Answer:

f1= -350cm or -3.5m

f2= 22.1cm or 0.221m

Explanation:

A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.

Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.

3 0

3 years ago

A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125

frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0

3 years ago