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Softa [21]
3 years ago
9

A coin dropped from the top of a precipice (cliff) takes 2.42 to hit the ground. How highis the precipice (cliff)? .

Physics
1 answer:
goldfiish [28.3K]3 years ago
4 0

Answer:

28.7 m.

Explanation:

Hello,

In this case, since the coin is dropped from the rest condition at which the initial velocity is 0 m/s, we can compute the height of the precipice via the following equation:

y=y_0+v_0*t-\frac{1}{2}gt^2

Whereas the final height is 0, the initial one is to be computed and the gravity is considered as 9.8 m/s², therefore the height turns out:

y=y_0+v_0*t-\frac{1}{2}gt^2\\\\y_0=\frac{1}{2}gt^2=\frac{1}{2}*9.8m/s^2*(2.42s)^2 \\\\y_0=28.7m

It means that the height of the cliff is 28.7 m.

Regards.

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Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How
grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (\Delta s), in meters, can be determined by the following expression:

\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2} (1)

Where:

v_{o} - Initial velocity, in meters per second.

t - Time, in seconds.

a - Acceleration, in meters per square second.

If we know that v_{o} = 5\,\frac{m}{s}, t = 6\,s and a = 2\,\frac{m}{s^{2}}, then the box displacement after 6 seconds is:

\Delta s = 66\,m

The box displacement after 6 seconds is 66 meters.

5 0
3 years ago
Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

8 0
3 years ago
HELP!!!! WILL MARK BRAINLIEST!!!! IF YOU LEAVE AN ANSWER EXPLAIN! THANKS
Travka [436]

Answer:

the answer is C

Explanation:

we know this because if you compare the graphs and look at the direction. it isn't always in the explanation or the few sentences they gave you at the top. also, look at the waves, you can see in Davids drawing that it is directly straight up, A and B do not represent that. A isn't even a valid answer. Notice also in A that the arrow is going in the completely different direction than in Davids drawing. B is also going a different direction even though it is only turned a little bit although if it was straight up like Davids drawing then it would most likely be a correct answer. C does have one arrow going a different direction but look at how it has two, showing in which if the waves were to turn then the arrow is still valid

7 0
3 years ago
Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s
vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances  

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3 years ago
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pashok25 [27]
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