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Cerrena [4.2K]
3 years ago
5

A student drops a rock from a bridge to the water 12m below. How many seconds does it take the rock to hit the water?

Physics
1 answer:
ANEK [815]3 years ago
7 0
R = \frac{g t^{2} }{2}
r - displacement or height
t - time taken
g = 10 m/s-2
t = square root from 2r/g = \sqrt{ \frac{2 x 12}{10} } = 1.5 seconds
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What conditions can create the largest waves in the ocean. this is for science. i need done asap.
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7 0
2 years ago
An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan
MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

Vertical \ velocity \ (\frac{m}{s} ) =  u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) =  u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) =  a_{v} =  9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) =  a_{h} =  0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...

7 0
2 years ago
Relationship between voltage, resistance and current in a circuit
Lena [83]

Answer:

V = I×R

where -

V = potential difference across

I = current flowing in the circuit

R = Equivalent Resistance in the circuit

8 0
2 years ago
1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Alik [6]

Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

Mass of a car, m = 1550 kg

Speed pf car, v = 15 m/s

We need to find the momentum of the car. The formula for the momentum of an object is given by :

p = mv

Substituting all the values in the above formula

p = 1550 kg × 15 m/s

p = 23250 kg m/s

So, the momentum of the car is 23250 kg m/s.

3 0
3 years ago
It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T
tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\

<u>ω = 0.05 rad/s</u>

7 0
3 years ago
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