I would say A Because it weighs more than the water
Answer:
Explanation:
As we know that Far sighted person has near point shifted to 80 cm distance
so he is able to see the object 80 cm
now the distance of lens from eye is 2 cm
and the person want to see the objects at distance 10 cm
so here the image distance from lens is 80 cm and the object distance from lens is 8 cm
now from lens formula we have
Answer:
y = y₀ (1 - ½ g y₀ / v²)
Explanation:
This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i
y = y₀ + v₀ t - ½ g t²
y = y₀ - ½ g t²
for the ball thrown from the ground with initial velocity v₀₂ = v
y₂ = y₀₂ + v₀₂ t - ½ g t²
in this case y₀ = 0
y₂2 = v t - ½ g t²
at the point where the two balls meet, they have the same height
y = y₂
y₀ - ½ g t² = vt - ½ g t²
y₀i = v t
t = y₀ / v
since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height
y = y₀ - ½ g t²
y = y₀ - ½ g (y₀ / v)²
y = y₀ - ½ g y₀² / v²
y = y₀ (1 - ½ g y₀ / v²)
with this expression we can find the meeting point of the two balls
The work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.
<h3>How to calculate work done?</h3>
Work done is a measure of energy expended in moving an object; most commonly, force times distance.
It is said that no work is done if the object does not move, hence, the work done on an object can be calculated as follows:
Work done = Force × Distance
According to this question, a student carries a very heavy backpack and to lift the backpack off the ground, the student must apply 80 N of force to lift the backpack 1.5 m.
Work done = 80N × 1.5m
Work done = 120J
Therefore, the work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.
Learn more about work done at: brainly.com/question/28172139
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Answer: The final mass of the tank is 2.46kg
Explanation: All shown in the attachment.
Assumptions:
i. Argon is treated as an ideal gas at the specified conditions.
ii. Isentropic relation of ideal gas applies at the given conditions.
