Use the Second Law of Newton, which states that Net Force = mass times acceleration
F = m * a
F = 50 kg * 1.5 m/s^2 = 75 N
Answer: option C
"Light year" is a distance, not a speed. It's the distance light travels in one year, at the speed of 299,792,458 meters per second.
Answer:
165.8 V/m
Explanation:
The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)
Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)
V = Q/C
V = Q ÷ 2πε₀L/㏑(b/a)
V = Q㏑(b/a)/2πε₀L
So, the potential difference per unit length V' is
V' = V/L = Q㏑(b/a)/2πε₀
Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have
V' = Q㏑(b/a)/2πε₀
V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)
V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)
V' = 0.16575 × 10³ V/m
V' = 165.75 V/m
V' ≅ 165.8 V/m
Assuming that you're given either an initial or final velocity, you can use the following equation and solve for the initial or final velocity.
Vyf² = Vyi² - 2g(y - y₀)
Where,
Vyf² = final velocity
Vyi₂² = initial velocity
g = 9.81 m/s²
(y - y₀) = the change in the distance along the y-axis.
You'll need also determine the positive and negative of your y-axis for your final solution because velocity can be positive or negative based on direction. Lastly, don't forget to square root both sides of your equation for your velocity.
I hope this helps.
Grav. Potential at surface of the asteroid:
V = - G.Ma./ R
V = (-) 6.67^-11 x 4.0^20kg / 5.7^5m .. .. V = (-) 4.681 *10 ^5 J/kg
The GPE of the package on the asteroid = 9.0kg x (-) 4.681*10^5J/kg = (-) 4.21 ^5
J
This is the amount of energy required to come back the
package to infinity.
The total energy that needs to be transported to the package:
GPE + KE(for 187m/s)
Total energy required E = 4.21*10^5 + (½x 9.0kg x 168²) = 5.48 * 10^5 J
When the required energy is to be complete by releasing a compressed spring,
Elastic PE stored in spring = ½.ke² = 5.48 * 10^5 J where e = compression
distance
e = √ (2 x 5.48*10^5 / 2.1*10^5)
e = 2.28 m
