PHYSICS
MIDDLE SCHOOL

Answer: Kinetic Energy = 1/2 Mass (kg) * velocity squared (m/s)

KE = 1/2mV^2

KE = 1/2(0.15 kg)(40 m/s)^2

The answer is 120 joules.

KE = 1/2mV^2

KE = 1/2(0.15 kg)(40 m/s)^2

The answer is 120 joules.

Answer:

0.5mv^2 > 0.5*0.15*40*40=120J

COLLEGE

The presence of which element would indicate that a star is going through a high-mass star life cycle as opposed to a low-mass star life cycle? hydrogen

Celestial bodies in the universe like the stars, gain their energy by nuclear fusion. This is a nuclear reaction that emits radiation by joining subatomic particles together to yield another new element. This cause by instability of certain elements due to their high neutron-to-proton ratio. The most stable element there is, is Fe-26. Elements lighter than Fe-26 are most likely to undergo nuclear fusion (combining), while elements heavier than Fe-26 are most likely to undergo nuclear fission (breaking).

So that is how the Sun gains its energy. It is very abundant in hydrogen, such that hydrogen undergoes nuclear fusion. Two protons from two hydrogen atoms combine at very very high temperatures to form a Helium atom. Therefore, a high-mass star life is very abundant in Hydrogen, while a low-mass star life is very abundant in Helium.

So that is how the Sun gains its energy. It is very abundant in hydrogen, such that hydrogen undergoes nuclear fusion. Two protons from two hydrogen atoms combine at very very high temperatures to form a Helium atom. Therefore, a high-mass star life is very abundant in Hydrogen, while a low-mass star life is very abundant in Helium.

COLLEGE

A concave lens has a focal length of 20 cm. A real object is 30 cm from the lens. Where is the image? What is the magnification?

Answer:

12 cm and 0.4

Explanation:

f = - 20 cm, u = - 30 cm

Let v be the position of image and m be the magnification.

Use lens equation

1 / f = 1 / v - 1 / u

- 1 / 20 = 1 / v + 1 / 30

1 / v = - 5 / 60

v = - 12 cm

m = v / u = - 12 / (-30) = 0.4

COLLEGE

A soccer ball starts from rest and accelerates with an acceleration of 0.395 m/s^2 while moving down a 8.50 m long inclined plane. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.75 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane (in m/s)? (Round your answer to at least two decimal places.) (b) How long does it take to roll down the first plane (in s)?

Answer:

a) The speed of the ball at the bottom of the first plane is 2.59 m/s

b) It takes the ball 6.56 s to roll down the first plane.

Explanation:

The equations for the position and velocity of the ball are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity at time t

b) First, let´s calculate the time it takes the ball to reach the bottom of the plane using the equation for the position:

x = x0 + v0 · t + 1/2 · a · t²

Placing the center of the frame of reference at the point where the ball starts rolling, x = 0. Since the ball starts from rest, v0 = 0. Then:

x = 1/2 · a ·t²

Let´s find the time when the ball reaches a position of 8.50 m

8.50 m = 1/2 · 0.395 m/s² · t²

t² = 2 · 8.50 m / 0.395 m/s²

t = 6.56 s

a) Now, using the equation of the velocity, we can calculate the velocity of the ball at the bottom of the plane (t = 6.56 s):

v = a · t

v = 0.395 m/s² · 6.56 s = 2.59 m/s

MIDDLE SCHOOL

What is the proper hand position when driving straight ahead

NHTSA (National Highway Traffic Safety Administration) now recommends the technique known as 9 and 3. Place your left hand on the left portion of the steering wheel in a location approximate to where the nine would be if the wheel was a clock. Your right hand should be placed on the right portion of the wheel where the three would be located.