Answer:
Q = 6.33μC
Explanation:
To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:
M: mass of the ball = 0.09kg
T: tension of the string
F_e: electric force between charges
angle = 45°
The electric force is given by:
Q: charge of the balls
r: distance between balls = 2m
You divide both equation in order to eliminate the tension T:
By doing Q the subject of the formula and replacing you obtain:
hence, the charge of the balls is 6.33μC
Answer:
After walking across a carpeted floor in socks, Jim brings his finger near a metal doorknob and receives a shock. What does this demonstrate? Frictional forces require direct contact. Electrical forces can act at a distance.
Answer:
1) R1 + ((R2 × R3)/(R2 + R3))
2) 0.5 A
3) 3.6 V
Explanation:
1) We can see that resistors R2 and R3 are in parallel.
Formula for sum of parallel resistors; 1/Rt = 1/R2 + 1/R3
Making Rt the subject gives;
Rt = (R2 × R3)/(R2 + R3)
Now, Resistor R1 is in series with this sum of R2 and R3. Thus;
Total resistance of circuit = R1 + ((R2 × R3)/(R2 + R3))
2) R_total = R1 + ((R2 × R3)/(R2 + R3))
We are given;
R1 = 7.2 Ω
R2 = 8 Ω
R3 = 12 Ω
R_total = 7.2 + ((8 × 12)/(8 + 12))
R_total = 7.2 + 4.8
R_total = 12 Ω
Formula for current is;
I = V/R
I = 6/12
I = 0.5 A
3) since current through the circuit is 0.5 and R1 is 7.2 Ω.
Thus, potential difference through R1 is;
V = IR = 0.5 × 7.2 = 3.6 V
Answer:
Gold
Explanation:
Given:
Mass of sample = 63.5 g
Mass of water = 60.2 g
Find:
Object
Computation:
Mass of water displaced = 63.5 g - 60.2 g
Mass of water displaced = 3.3 g
So, volume in water = 3.3 cm³
Density = Mass / Volume
Density = 63.5 g / 3.3
Density = 19.24
So,
Object ,must be gold.
