Answer:
1275J
Explanation:
Given parameters:
Force on box = 85N
Distance moved = 15m
Unknown:
Work done = ?
Solution:
Work done is the amount of force applied on a body to move it through a specific distance.
Work done = Force x distance
Now insert the parameters and solve;
Work done = 85 x 15 = 1275J
Answer:
Geology is the study of the Earth that involves the process at Earth, materials of which it is made, and its history.
<u>Geologists combine both laboratory and field data to illustrate the results of their research. Some observations that can the geologist make by working outdoors instead of in a lab are as follows:</u>
- Understanding and exploring the earth's surface closely using geophysical tools.
- Collecting samples by own and make some interpretations at the same time.
- Observation of the landscapes
- Close observation of outcrops
Answer:
1. 24375 N/C
2. 2925 V
Explanation:
d = 12 cm = 0.12 m
F = 3.9 x 10^-15 N
q = 1.6 x 10^-19 C
1. The relation between the electric field and the charge is given by
F = q E
So, 
E = 24375 N/C
2. The potential difference and the electric field is related by the given relation.
V = E x d
where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.
By substituting the values, we get
V = 24375 x 0.12 = 2925 Volt
Because of symmetry electric field component in the x axis cancels out. Now just use electric field formula and slap that sine of theta cause you want the vertical component of electric field and multiply that by two since there’s two charges. I’ve shown my work. Hope it helps✌
Answer:
a)W= - 720 J
b)ΔU= 330 J
Explanation:
Given that
P = 0.8 atm
We know that 1 atm = 100 KPa
P = 80 KPa
V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)
V₂ = 3 L = 0.003 m³
Q= - 390 J ( heat is leaving from the system )
We know that work done by gas given as
W = P (V₂ -V₁ )
W= 80 x ( 0.003 - 0.012 ) KJ
W= - 0.72 KJ
W= - 720 J ( Negative sign indicates work done on the gas)
From first law of thermodynamics
Q = W + ΔU
ΔU=Change in the internal energy
Now by putting the values
- 390 = - 720 + ΔU
ΔU= 720 - 390 J
ΔU= 330 J
