Answer:
work done lifting the bucket (sand and rope) to the top of the building,
W=67.46 Nm
Explanation:
in this question we have given
mass of bucket=20kg
mass of rope=
height of building= 15 meter
We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand
work done in lifting the rope is given as,
=
..............(1)
=
=22.5 Nm
work done in lifting the sand is given as,
.................(2)
Here,
F=mx+c
here,
c=20-18
c=2
m=
m=.133
Therefore,
Put value of F in equation 2
Therefore,
work done lifting the bucket (sand and rope) to the top of the building,
W=22.5 Nm+44.96 Nm
W=67.46 Nm
Energy Density = 1/2 × ε(0) × (V/d)^2
V = 100, d = 0.01, ε(0) = 8.85 x 10^-12
Answer:
there should some picture to identify right?
pls edit your question and insert the picture..
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
The given mass is 0.025563 g.
Examine the given choices.
a. 0.026 g
This uses 2 significant digits, with rounding to the 3rd decimal place.
b. 2.5 x 10² g = 250 g.
It is incorrect.
c. 0.025 g.
This uses 2 significant digits. It is inaccurate because it does not use rounding to the 3rd decimal place.
d. 0.02 g
This uses one significant digit. It is incorrect for representing the given data.
Answer: a. 0.026 g
