6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
7. Same idea as question 2.
First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N.
10. First find the force of block B to the right due to its acceleration:
F = ma
F = (24)(0.5)
F = 12N
So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:
The sum of the forces in the leftward and rightward direction for block B must equal 12N.
75 - x = 12
x = 63N
So the force of friction of block A on block B is 63N to the left.
If the object is kept in between the principle axis and the focus but some what nearer to the focus then we will get the enlarge,erect,and real image.
Answer:
See explanation
Explanation:
The question is incomplete because the images were not attached but I will try to help you as much as possible.
Constant acceleration implies that the velocity increases uniformly with time.
The graph of constant acceleration is a straight line graph having a slope. The slope of the graph is constant at any point along the straight line.
The image attached shows a velocity-time graph depicting constant acceleration.
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
Actually, the real reason it appears yellow-white is because of the temperature of the star. the stars that are hotter tend to be more blue and the stars that aren’t as hot tend to be more red/orange/yellow.
