Answer:
1.6s
Explanation:
Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,
To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.
Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh
Where h = 5.2sin25
h = 2.2 m
P.E = 1.2 × 9.8 × 2.2
P.E = 25.84 j
At the bottom, K.E = P.E
1/2mv^2 = 25.84
Substitutes mass into the formula
1.2 × V^2 = 51.69
V^2 = 51.69/1.2
V^2 = 43.07
V = 6.56 m/s
Using the third equation of motion
V^2 = U^2 + 2as
Since the object started from rest,
U = 0
6.56^2 = 2 × a × 5.2
43.07 = 10.4a
a = 43.07/10.4
a = 4.14 m/s^2
Using the first equation of motion,
V = U + at
Where U = 0
6.56 = 4.14t
t = 6.56/4.14
t = 1.58s
Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s
Answer:
The speed of the wave is 15ms .
Explanation:
Kinetic energy = mass time squared speed divided by 2
W=mv^2/2 = 50*10*10/2 = 2500 J
Net force = Pull force - Friction (Opposing force) = 700 - 215 = 485 N
Given parameters:
Initial volume = 75cm³
New volume = 30cm³
New pressure = 110Pa
Unknown:
Initial pressure = ?
Solution:
Condition: temperature is constant
We simply apply Boyle's law to this problem:
" the volume of a fixed mass(mole) of a gas varies inversely as the pressure changes if the temperature is constant".
Mathematically;
P₁ V₁ = P₂ V₂
where P and V are pressure and volume values
1 and 2 are the initial and final states.
Input the parameters and solve for P₁;
P₁ x 75 = 110 x 30
P₁ = 44Pa
The initial pressure is 44Pa
