You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.
T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N
Explanation:
since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.
We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.
range can be calculated by the formula :-
u is the velocity during its take off and 
is the angle at which its thrown
Given that
- u = 8m/ s
- = 40°
calculating range using the above formula
value of sin 80 = 0. 985
Hence,
An object has undergone acceleration if ...
-- it's moving faster than it was before
or
-- it's moving slower than it was before
or
-- it's moving in a different direction that it was before.
Answer:
14.49 g/cm²
Explanation:
I = Io e^-(ux)
Where:
I = 573
Io = 1045
x = 0.3 inches and
rho = 11.4g/cm^3
Using the conversion constant
1 inch = 2.54 cm;
0.3 inches = 0.3 * 2.54 cm
0.3 inches = 0.762 cm
I/Io = e^-(ux), or say
Io/I = e^(ux), taking the In of both sides
ln(Io/I) = ux, making u subject of formula
u = 1/x * ln(Io/I)
u = 1/0.762 * ln(1045/573)
u = 1.312 * 0.6
u = 0.787
Next, we say that
u/rho = 0.7872/11.4 = 0.069
And finally, we make
1/(u/rho) to be our final answer
Inverse of the answer is = 14.49 g/cm²
Therefore, the um^-1 in g/cm^2? is 14.49
