Speed v = initial speed u + acceleration a x time t
v=u+at = 2 + 4*3 = 14 m/s
12.00 min = 0.2 hr
8.00 min = 0.15 hr
Total distance:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)
= 8.25 km
Average speed:
(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3
= 15 km/hr
Change in position:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)
= 0.25 km
Average velocity:
(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3
≈ 1.67 m/s
Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases
The relation between potential difference and the electric field is given by ΔV = E.d
Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.
The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.
The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.
Refer to more about the potential difference here
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Answer:
Explanation:
position of centre of mass of door from surface of water
= 10 + 1.1 / 2
= 10.55 m
Pressure on centre of mass
atmospheric pressure + pressure due to water column
10 ⁵ + hdg
= 10⁵ + 10.55 x 1000 x 9.8
= 2.0339 x 10⁵ Pa
the net force acting on the door (normal to its surface)
= pressure at the centre x area of the door
= .9 x 1.1 x 2.0339 x 10⁵
= 2.01356 x 10⁵ N
pressure centre will be at 10.55 m below the surface.
When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .
Answer:
Answered
Explanation:
x= 0.02 m
E_p= 10.0 J
E_p= 0.5kx^2
10= 0.5k(0.02)^2
solving we get
K= 50.0 N/m
Now
E'_p= 0.5kx'^2
E'_p= 0.5×50×(0.04)^2
E'_p=40 J
b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both
c) 20 J = 0.5×50,000×x^2
solving
x= 0.028 m
d) k is 50.0 N/m from above calculation
