Answer:
22.14 m/s
Explanation:
height, h = 25 m
diameter, = 1.2 cm
Let v be the velocity of water.
v = 22.14 m/s
Thus, the speed of water is 22.14 m/s .
M1 descending
−m1g + T = m1a
m2 ascending
m2g − T = m2a
this gives :
(m2 − m1)g = (m1 + m2)a
a =
(m2 − m1)g/m1 + m2
= (5.60 − 2)/(2 + 5.60) x 9.81
= = 4.65m/s^2
Answer:
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine (
)
Explanation:
Substituting values of r, ρ, σ, v & g in the equation:
Answer:
A. a set of mathematically topics that are relevant to introductory physics.
Explanation:
The physics primer is not defined as the online comprehensive mathematics textbooks. It is the set of topics of mathematics which gives students trouble and remember.
Therefore, it is defined as the process of physics problem solving. So, mathematically skills are covered in physics course as a primer related success.
Therefore, it is a set of topics of mathematics that are relevent to introductory physics.
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
</span>
<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
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