Answer:
7/150
Explanation:
The following data were obtained from the question:
Object distance (u) = 75cm
Image distance (v) = 3.5cm
Magnification (M) =..?
Magnification is simply defined as:
Magnification (M) = Image distance (v)/ object distance (u)
M = v /u
With the above formula, we can obtain the magnification of the image as follow:
M = v/u
M = 3.5/75
M = 7/150
Therefore, the magnification of the image is 7/150.
Answer:
(B) 0.5 g
Explanation:
Newton's second law says ∑ F i = m a .
the rate of change in momentum of a body is proportional to the force applied on the body.
f∝ma
f=kma
were k is constant and equal to 1
The centripetal acceleration is an acceleration.
the tension on the swing and object weight goes to the left hand side while the centripetal acceleration goes to the right handside
At the bottom of the swing, ΣF = FT – mg = mac;
notice that the tension in the swing is 1.5 times the weight of the object
we can write
1.5mg – mg = mac,
0.5mg = mac
0.5 g=ac
Answer:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Explanation:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
The modern name, Mount St. Helen's, was given to the volcanic peak in 1792 by seafarer and explorer Captain George Vancouver of the British Royal Navy. He named it in honor of fellow countryman Alleyne Fitzherbert, who held the title 'Baron St. Helen's.
