The height at which the mass will be lifted is; 3 meters
<h3>How to utilize efficiency of a machine?</h3>
Formula for efficiency is;
η = useful output energy/input energy
We are given
η = 60% = 0.6
Input energy = 4 KJ = 4000 J
Thus;
0.6 = useful output energy/4000
useful output energy = 0.6 * 4000
useful output energy = 2400 J
Work done in lifting mass(useful output energy) = force * distance moved
Useful output energy = 800 * h
where h is height to lift mass
Thus;
800h = 2400
h = 2400/800
h = 3 meters
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True.
To understand it better
First job : Pet shop
Second job : pizza place
The first job supports his career path he has experience.
The second job support life in making sure he gets to his career path/ does help financially for him to get there.
And it’s called career pathway.
Answer:
Environmental
Explanation:
weather , noise, heat ect
Answer:
P = 18035.25 N
Explanation:
Given
D = 10.4 mm
ΔD = 3.2 ×10⁻³ mm
E = 207 GPa
ν = 0.30
If
σ = P/A
A = 0.25*π*D²
σ = E*εx
ν = - εz / εx
εz = ΔD / D
We can get εx as follows
εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴
Now we find εx
ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³
then
σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa
we have to obtain A:
A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²
Finally we apply the following equation in order o get P
σ = P/A ⇒ P = σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N
Answer:
Explanation:
Initial speed of the mobile = 3.2 m/s
Final speed of the mobile = 5.2 m/s
Time, t = 8 s
We need to find the acceleration of the mobile. It can be given by the change in velocity divided by time. So,
So, the acceleration of the mobile is 
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