Answer:
The tension in the string at that point is 90.75 N
Explanation:
Given;
mass of the object, m = 3 kg
length of string, r = 0.25 m
the angular velocity, ω = 11 rad/s
The tension on string can be equated to the centrifugal force on the object;
T = mω²r
Where;
T is the tension in the string
m is mass of the object
ω is the angular velocity
r is the radius of the circular path
T = 3 x (11)² x 0.25
T = 90.75 N
Therefore, the tension in the string at that point is 90.75 N
Answer:
Value of each charge is given as
Explanation:
As we know that electrostatic force between two charges is given as
here we know that
r = 0.100 m
now we have
Answer:
B
Explanation:
In this calorimetry problem, the heat released by the reaction is equal to the heat absorbed by the solution (assumed to have the same specific heat capacity as water, 4.19 Jg⁻¹°C⁻¹).
The formula Q = mcΔt will be used to calculate the heat energy, where m is the mass, c is the specific heat capacity, and Δt is the change in temperature from final to initial.
The volume of solution is (50.0 + 50.0)mL = 100.0mL = 100.0g, since water has a density of 1.00g/mL.
The heat absorbed by the solution is then calculated.
Q = mcΔt = (100.0 g)(4.19 Jg⁻¹°C⁻¹)(28.2°C - 25.0°C) = 1340 J
The closest answer is B) 1300 J. This answer is obtained by including only two significant figures in the answer.
Answer:
d = 11.1 m
Explanation:
Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:
Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:
Plugging this into the energy conservation equation and cancelling m, we get
Solving for d,
It is called a Nanomaterial