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3 years ago

A 10 kg rock is suspended 320 m above the ground.

1 answer:

7 0

Potential energy is energy which results from position or configuration. It is the energy stored in an object not moving. It is calculated by:

PE = mgh

where m is the mass, g is the gravitational acceleratin and h is the height.

PE = 10 ( 9.8) (320) = 31360 J = 3.1 x 104 J --------> OPTION 2

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Moore's Law postulates that:

iris [78.8K]

Answer:

Option a. the number of transistors per square inch on integrated circuits will double every two years.

Explanation:

Moore's law states that the number of transistors in an integrated circuit will double every two years.

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3 years ago

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3 years ago

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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th

melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore, the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F = 624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

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3 years ago

If a galaxy has an apparent radial velocity of 2000 km/s and the Hubble constant is 70 km/s/Mpc, how far away is the galaxy

ddd [48]

Answer:

28.57 Mpc

Explanation:

This question is going to be solved by applying Hubble's Law.

This Hubble's Law is actually an observation in physical cosmology. This observation makes it clear that galaxies are moving away from the Earth, and are doing so at speeds proportional to their distance. This essentially means that the farther they are from the Earth, the faster they are moving away from Earth.

It is represented by this formula

v = H(0)D, where

v = speed

H(0) = Constant of proportionality, or otherwise, Hubble's constant.

D = Distance to a galaxy

Applying the given parameters to the formula, we have

v = H(0).D

D = v / H(0)

D = 2000 / 70

D = 28.57 Mpc

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3 years ago

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

kherson [118]

Answer:

a. The rock takes 2.02 seconds to hit the ground

b. The rock lands at 20,2 m from the base of the cliff

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

$\displaystyle t=\sqrt{\frac{2h}{g}}$