Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps
instantaneous speed of the car
Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
<h3>
Acceleration of the cart</h3>
The acceleration of the cart is determined from the net force acting on the mass-cart system.
Upward force = Downward force
ma = mg
13a = 10(9.8)
13a = 98
a = 98/13
a = 7.54 m/s²
Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
Learn more about acceleration here: brainly.com/question/14344386
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