Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151
Answer:
563712.04903 Pa
Explanation:
m = Mass of material = 3.3 kg
r = Radius of sphere = 1.25 m
v = Volume of balloon = 
M = Molar mass of helium = 
= Density of surrounding air = 
R = Gas constant = 8.314 J/mol K
T = Temperature = 345 K
Weight of balloon + Weight of helium = Weight of air displaced

Mass of helium is 6.4356 kg
Moles of helium

Ideal gas law

The absolute pressure of the Helium gas is 563712.04903 Pa
Answer:
A spring whose spring constant is 200 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required to compress it another 1 inch.
Step 1 of 4
The force at any point during the deflection of the spring is given by,
where is the initial force
and x is the deflection as measured from the point where the initial force occurred.
The work required to compress the spring is
Therefore work required to compress the spring is
The work required to compress the spring in Btu is calculated by
Where 1Btu =778
The work required to compress the spring,
eman Asked on February 19, 2018 in thermal fluid Sciences 4th solutions.
Explanation:
Yes, they live off of other organisms and harm the organisms.
Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force 
Mass of object = 2.00 kg
Initial position 
Final position 
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done


Put the value into the formula




Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.