Answer:
Subduction , Latin for "carried under," is a term used for a specific type of plate interaction. It happens when one lithospheric plate meets another—that is, in convergent zones —and the denser plate sinks down into the mantle.
Is this right or they wrong definitions which ones are the right ones someone !!!!!
Answer:
They are right.
Explanation:
Answer:
Mechanical Energy : KE + PE
Conversion : "When energy transfers from one form to another"
Potential Energy: the energy possessed by a body by virtue of its position relative to others , stresses within itself, electric charge , and other factors .'
Kinetic Energy: energy of an object in motion
Law of conservation of energy: KE+PE+friction=KE
Explanation:
First of all mechanical energy is kinetic energy plus potential energy (it is the energy of movement) So:
Mechanical Energy : KE + PE
Conversion is when energy converts or becomes a different form. So:
Conversion : "When energy transfers from one form to another"
Potential energy is stored energy, in Physics I or AP Physics I, it is often due to it being at a height, but batteries, foods, etc. are also example of it, so:
Potential Energy: the energy possessed by a body by virtue of its position relative to others , stresses within itself, electric charge , and other factors .'
Kinetic energy is for objects in motion so you got it right!
Kinetic Energy: energy of an object in motion
The law of conservation of energy means there is the same amount of energy before, as there is after, so when you see an equation with energy on both sides, it is usually this. Also, this is the last question left, so this has to be the answer.
Law of conservation of energy: KE+PE+friction=KE
Which of the following is true of the deep
water layer of the ocean?
A. warmest and least dense of the ocean layers
B. experiences a rapid decrease in temperature
C. is warm in the summer and cold in the winter
D. cold all year round
Brian Lara is a cricketer playing in the field on the second day of a cricket test-match. He exerts a forward force on the 0.145kg cricket ball, as he catches it, to bring it to rest from a speed of 38.2m/s. During the process, his hand recoils a distance of 0.135m. Determine the acceleration of the ball and the force which is applied to it by Brian Lara.
Answer:
a = -3984.6 m/s²
F = 577.76 N
Explanation:
The acceleration of the ball can be calculated by using the third equation of motion:
[tex]2as = v_f^2 - v_i^2\\[/tex]
where,
a = acceleration of ball = ?
s = distance covered = recoil distance = 0.135 m
vf = final speed = 0 m/s
vi = initial speed = 38.2 m/s
Therefore,
[tex]2(0.135\ m)a = (0\ m/s)^2-(38.2\ m/s)^2\\[/tex]
a = -3984.6 m/s²
here negative sign shows deceleration.
Now, for the force applied by Brian Lara will be equal in magnitude but opposite in direction of the force required to stop the ball:
[tex]F = -ma\\F = -(0.145\ kg)(-3984.6\ m/s^2)\\[/tex]
F = 577.76 N
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. Bifocal lenses are used to correct his vision. A diverging lens is used in the top part of the lens to allow the person to clearly see distant objects.
1. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
2. What power lens (in diopters) should be used in the bottom half of the lens to allow him to clearly see objects 25 cm away?
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
[tex]\frac{1}{f_1} = \frac{1}{q}[/tex]
q = f₁
2) for an object located at p = 25 cm
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]
[tex]\frac{1}{f_2}[/tex] = 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = [tex]\frac{1}{f}[/tex]
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
3.
Two Cars, A and B, (starting, at the same time, from the same point) are moving
with average speeds of 40 km/h and 50 km/h, respectively, in the same direction.
Find how far will Car B be from Car A after 3 hours.
Answer:
car B will be 30 Km ahead of car A.
Explanation:
We'll begin by calculating the distance travelled by each car. This is illustrated below:
For car A:
Speed = 40 km/h
Time = 3 hours
Distance =?
Speed = distance / time
40 = distance / 3
Cross multiply
Distance = 40 × 3
Distance = 120 Km
For car B:
Speed = 50 km/h
Time = 3 hours
Distance =?
Speed = distance / time
50 = distance / 3
Cross multiply
Distance = 50 × 3
Distance = 150 Km
Finally, we shall determine the distance between car B an car A. This can be obtained as follow:
Distance travelled by car B (D₆) = 150 Km
Distance travelled by car A (Dₐ) = 120 Km
Distance apart =?
Distance apart = D₆ – Dₐ
Distance apart = 150 – 120
Distance apart = 30 Km
Therefore, car B will be 30 Km ahead of car A.
