Answer:
r₁ = 20.5 cm
Explanation:
In this exercise we can use the conservation of energy
the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign
starting point.
Em₀ = U_g + U_e + K =
the two in the kinetic energy is because they are two particles
final point. When it is detained
Em_f = U_g + U_e =
the energy is conserved
Em₀ = em_f
the charges and masses of the two particles are equal
sustitute the values
-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1
-5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁ + 8.1 10⁻⁶ / r₁
We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.
3.24 10⁻⁵ - 7.2 10⁻⁵ = 8.1 10⁻⁶ / r₁
-3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁
r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵
r₁ = 2.045 10⁻¹ m
r₁ = 20.5 cm
No, the car travels 1 metre in 5s at the start which is 0.2m/s, while the second meter it travels one metre in 8 seconds which is 0.125 m/s, the speed changes therefore it is not constant during the two metres the car travels
Potassium iron magnesium aluminum silicate hydroxide and Potassium magnesium aluminum silicate hydroxide.
Some types of storage tanks are exempt from the federal regulations on USTs, including
: A) tanks that store heating oil burned on the premises.
<h3>What are Underground Storage Tank Systems?</h3>
Underground Storage Tank systems are tanks along with the underground piping that has at least 10% of their total volume underground.
Some of the storage tanks that are exempt from the federal regulations on USTs are;
- Farm and residential tanks of 1,100 gallons or less capacity holding motor fuel used for noncommercial purposes.
- Tanks storing heating oil used on the premises where it is stored.
- Tanks of 110 gallons or less capacity.
Learn more about Underground Storage Tank systems here:
brainly.com/question/26571
Given that,
Mass = 10 kg
Angle = 30°
Static friction = 0.582
Kinetic friction = 0.528
We need to calculate the parallel force
Using balance equation

Put the value into the formula


Let’s determine the two friction forces
We need to calculate the static friction force
Using static friction force

Put the value into the formula


Since the box is already moving, the force of the push must be greater than difference of the force parallel and the static friction force.
We need to calculate the kinetic friction force
Using kinetic friction force

Put the value into the formula


Since this friction force is less than 49 N, the box will accelerate as it moves down hill.
Hence, This is required solution.