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2 years ago

What is the final speed of a 60 kg boulder dropped from a 111 meter cliff

1 answer:

3 0

After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is

Square root of ( 19.6 M ) .

If M=111 meters, then the speed is <em>46.64 meters per second</em>.

We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.

You might be interested in

The four fundamental types of organic molecules are nucleic acids, proteins, lipids, and ________.

notka56 [123]

The other of the four organic molecules is carbohydrates.
All four of them are:
Nucleic Acids (DNA and RNA stuff)
Proteins
Lipids
& Carbohydrates

4 0

3 years ago

Read 2 more answers

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

6 0

2 years ago

A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work i

podryga [215]

The answer is 3) 480 joules

6 0

2 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

3 years ago

JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

3 0

3 years ago