The question is incomplete. The complete question is :
Two conducting spheres are mounted on insulating rods. They both carry some initial electric charge, and are far from any other charge. Their charges are measured. Then, the spheres are allowed to briefly touch, and the charge in one of them (sphere A) is measured again. These are the measured values:
a). Before contact:
Sphere A = 4.8 nC
Sphere B = 0 nC
What is the charge on sphere B after contact, in nC?
b). Before contact:
Sphere A = 2.9 nC
Sphere B = -4.4 nC
What is the charge on sphere B after contact, in nC?
Solution :
It is given that there are two spheres that are conducting and are mounted on an insulating rods which carry a initial charge and they are briefly touched and then one of the charge is measured.
Here the charge becomes divided when both the spheres are connected and then removed.
a). charge after they are charged
= 2.4 nC
b). The charge is
= -0.75 nC
Answer:
New volume (V2) = 60 liter
Explanation:
Given:
Amount of helium (V1) = 20 Liter
Temperature (T1) = 100°K
Temperature (T2) = 300°K
Find:
New volume (V2)
Computation:
According to Gas Law:
V1 / T1 = V2 / T2
20 / 100°K = V2 / 300°K
V2 = 60 liter
New volume (V2) = 60 liter
v₀ = initial velocity of the mobile = 10 m/s
v = final velocity of the mobile = 20 m/s
a = acceleration of the mobile = 5 m/s²
d = distance traveled during this operation = ?
Using the kinematics equation
v² = v²₀ + 2 a d
inserting the above values in the equation
20² = 10² + 2 (5) d
400 = 100 + 10 d
subtracting 100 both side
400 - 100 = 100 - 100 + 10 d
300 = 10 d
dividing both side by 10
300/10 = 10 d/10
d = 30 m
hence mobile travels 30 m.
<span>s=2.7 centimeters = 0.027 meters
t=3.9 milliseconds = 0.0039 seconds
s=(1/2)a*t^2
so
a=(2.7*2)/(0.0039)^2
= 355,029.58 m/s^2
a=355,029.58 m/s^2 = 355.02958 km / s^2</span>
Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
