Answer:
Explanation:
The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);
where 
is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.
Given;
v = 0.8m/s
g = 
r = ?
In order to solve for , we can simply make it the subject of formula from equation (1) as follows;
since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.
Therefore;
Answer:
10 m/s^2
Explanation:
Equation: F = ma.
a = acceleration
m = mass
F = force
Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.
F = 20
m = 2
a = ?
a = F/m
a = 20/2
a = 10 m/s^2
Answer: The temperature after another 5 minutes is 68.5°c
Explanation: Please see the attachments below
Based on the formula for calculating impulse, the impulse of the speed bump to the car is 2500 Ns.
<h3>
What is the impulse of the speed bump?</h3>
- Impulse = change in momentum
- momentum = mass * velocity
Change in momentum = mu - mv
where u is initial velocity
v is final velocity
Impulse = 500 * 20 - 500 * 15
Impulse = 2500 Ns
Therefore, the impulse of the speed bump to the car is 2500 Ns.
Learn more about impulse at: brainly.com/question/297527
Answer:
5.634 N rightwards
Explanation:
qo = - 3 x 10^-7 C
q1 = - 9 x 10^-6 C
q2 = 10 x 10^-6 C
r1 = 7 cm = 0.07 m
r2 = 20 cm = 0.2 m
The force on test charge due to q1 is F1 which is acting towards right
According to the Coulomb's law
F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)
F1 = 4.959 N rightwards
The force on test charge due to q2 is F1 which is acting towards right
According to the Coulomb's law
F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)
F2 = 0.675 N rightwards
Net force on the test charge
F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards
