(a)
The formula is:
∑ F = Weight + T = mass * acceleration
as the elevator and lamp are moving downward, I choose downward forces to be
positive.
Weight is pulling down = +(9.8 * mass)
Tension is pulling up, so T = -63
Acceleration is upward = -1.7 m/s^2
(9.8 * mass) + -63 = mass * -1.7
Add +63 to both sides
Add (mass * 1.7) to both sides
(9.8 * mass) + (mass * 1.7) = 63
11.5 * mass = 63
mass = 63 / 11.5
Mass = 5.48 kg
(b)
Since the elevator and lamp are going upward, I choose upward forces to be
positive.
Weight is pulling down = -(9.8 * 5.48) = -53.70
Acceleration is upward, so acceleration = +1.7
-53.70 + T = 5.48 * 1.7
T = 53.70 + 9.316 = approx 63 N
The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated
upward at the same rate of 1.7 m/s^2
The work done is 200 J
Explanation:
The work done by a force applied to move an object is given by:

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
In this problem, assuming that the force applied by the workers is parallel to the direction of motion of the crate, we have:
F = 10.0 N
d = 20.0 m

Therefore, the work done is:

Learn more about work here:
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Answer:
please read the answer below
Explanation:
The angular momentum is given by

By taking into account the angles between the vectors r and v in each case we obtain:
a)
v=(2,0)
r=(0,1)
angle = 90°

b)
r=(0,-1)
angle = 90°

c)
r=(1,0)
angle = 0°
r and v are parallel
L = 0kgm/s
d)
r=(-1,0)
angle = 180°
r and v are parallel
L = 0kgm/s
e)
r=(1,1)
angle = 45°

f)
r=(-1,1)
angle = 45°
the same as e):
L = 5kgm/s
g)
r=(-1,-1)
angle = 135°

h)
r=(1,-1)
angle = 135°
the same as g):
L = 5kgm/s
hope this helps!!
There are two every year – in September and March – when the Sun shines directly on the Equator and the length of day and night is nearly equal.
Answer:

Explanation:
Use the velocity formula to solve

In this question, you are given velocity
, and you are given a distance,
. Time in this question is what you'll need to find.
Start by rearranging the velocity formula, to isolate for t.

Start by multiplying both sides by t

Then divide both sides by v.

Now that you've isolated for time, sub in your values and calculate.
