As the ball is moving in air as well as we have to neglect the friction force on it
So we can say that ball is having only one force on it that is gravitational force
So the force on the ball must have to be represented by gravitational force and that must be vertically downwards
So the correct FBD will contain only one force and that force must be vertically downwards
So here correct answer must be
<em>Diagram A shows a box with a downward arrow. </em>
Answer:
71.8 N
Explanation:
T = Tension force in the strap
W = net work done = 752 J
f = force of friction = 8 N
d = displacement = 15 m
θ = angle between tension force and horizontal displacement = 36 deg
work done by frictional force is given as
W' = - f d
Work done by the tension force is given as
W'' = T d Cos36
Net work done is given as
W = W' + W''
W = T d Cos36 - f d
752 = T (15) Cos36 - (8) (15)
T = 71.8 N
Answer:
D. You will feel the same weight as you do on Earth
Explanation:
In free space, she is suppose to be weightless.
Free fall can be described as body in motion where the body is under the effect of acceleration due to gravity only and no other acceleration..
Since I am accelerating away from her at an acceleration of 1g
Then,
F=ma, where a=g
Then F=mg
Since my weight on earth is W=mg
This is equals to my weight in the spaceship, then I will feel the same weights as I do on earth.
Answer:
Density
Explanation:
Each of the three liquids have different densities. Have you ever tried to mix oil and water? They don't mix because they have different densities.
Answer:
a) F₁₂₀ = 1.34 pa A , b) F₂₀ = 0.746 pa A
Explanation:
Part. A
. The definition of pressure is
P = F / A
As the air can approach an ideal gas we can use the ideal gas equation
P V = n R T
Let's write this equation for two temperatures
P₁ V = n R T₁
P₂2 V = n R T₂
P₁ / P₂ = T₁ / T₂
point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂
P₂ = P₁ T₂ / T₁
P₂ = pa 393/293
P₂ = 1.34 pa
We calculate the strength
P₂ = F₁₂₀ / A
F₁₂₀ = 1.34 pa A
Part B
In this case the data is
Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure
P₁ / P₂ = T₁ / T₂
P₂ = P₁ T₂ / T₁
P₂ = pa 293/393
P₂ = 0.746 pa
Let's calculate the force (F20), from this point
F₂₀ / A = 0.746 pa
F₂₀ = 0.746 pa A