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Answer: 1.653 × 10⁴ N/C
Explanation: Given Q= 3.6× 10^-6
we assume there is a charge q placed at a distance r = 1.4m exerting an electric field strength towards Q.
Hence
F = kQq/r²
K= 9× 10^9
Now recall that E = F/q (electric field due to that charge)
E =( kQq/r²)/q
E = kQ/r² substituting the values gave the answer above.
Answer:
7167.27N/m
Explanation:
To get the force constant k, we will use the formula;
F = kx where
F is the force acting on the object down the plane
k is the spring constant
x is the extension
The force acting down the ramp us F = Wsintheta. The formula will become;
Wsintheta = kx
mgsintheta = kx
K = mgsintheta/x
K = (800)(9.8)sin47°/0.8
K= 7840sin47°/0.8
K = 5733.81/0.8
K = 7167.27N/m
Hence the spring constant of the spring is 7167.27N/m
1.47x10^5 Joules
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.