U = 0, initial vertical velocity
Neglect air resistance, and g = 9.8 m/s².
The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s
Answer: 2.0 s
Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:
diameter = 40 mm ==> area = <em>π</em> (0.04 m)² ≈ 0.00503 m²
diameter = 120 mm ==> area = <em>π</em> (0.12 m)² ≈ 0.0452 m²
Then the speed at the end of the 40 mm pipe is
(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s
(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s
Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Let m = mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.
Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)
The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
= [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
= 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)
Answer: 1.1x10¹⁶ kg
It would be water because if you freeze it than you will still be able to see it and if you boil it than you will be able to see it disappear.