Answer:
0.214 m
Explanation:
In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.
Therefore, we can write:

where
k is the Coulomb constant
is the charge on the balloon
is the charge on the bag
r is the separation betwen the bag and the balloon
is the mass of the bag
is the acceleration due to gravity
Solving for r, we find the distance at which the bag must be held:

Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as


R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as


i = 3.4 A
(c)
Current density is given as


J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as


E = 9.01 x 10⁻³ V/m
Answer:
514 cal
Explanation:
In order to calculate the lost heat by the amount of water you first take into account the following formula:
(1)
Q: heat lost by the amount of water = ?
m: mass of the water
c: specific heat of water = 1cal/g°C
T2: final temperature of water = 11°C
T1: initial temperature = 12°C
The amount of water is calculated by using the information about the density of water (1g/ml):

Then, you replace the values of all parameters in the equation (1):

The amount of water losses a heat of 514 cal
Answer:
T= 224.01 N
Explanation:
in imminent motion we have to :
- The frictional force reaches its maximum value
- The system is in balance of forces
Data
W= 500 N : weight of the log
μs = 0.5
μk = 0.35
α = 30°above the ground : angle of the cable attached to the log
Newton's first law to the log:
∑F =0 Formula (1)
∑F : algebraic sum of the forces in Newton (N)
Forces acting on the log
T: cable tension for impending movement
N: normal force
W : weight
f: frictional force , f= μsN
We apply the formula (1)
∑Fx=0
Tx-f = 0
Tcosα-μsN=0
Tcos30°-0.5N=0 Equation (1)
∑Fy=0
N+Ty-W=0
N+Tsin30°-500=0
N= 500-Tsin30° Equation (2)
We replace the value of N of the Equation (2) in the equation (1)
Tcos30°-0.5(500-Tsin30°) = 0
Tcos30°+0.5Tsin30° = 0.5*500
T( cos30°+0.5*sin30°) = 250
(1.116) T = 250
T= 250/1.116
T= 224.01 N
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