Any one trial might have been done incorrectly.
That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)
Hope this helps! (If correct, please rank as brainliest answer) :)
Answer:
Explanation:
Given that,
Bathysphere radius
r = 1.5m
Mass of bathysphere
M = 1.2 × 10⁴ kg
Constant speed of descending.
v = 1.2m/s
Resistive force
Fr = 1100N upward direction
Density of water
ρ = 1.03 × 10³kg/m³
The volume of the bathysphere can be calculated using
V = 4πr³ / 3
V = 4π × 1.5³ / 3
V = 14.14 m³
The Bouyant force can be calculated using
Fb = ρgV
Fb = 1.03 × 10³ × 9.81 × 14.14
Fb = 142,846.18 N
Buoyant force is acting upward
Weight of the bathysphere
W = mg
W = 1.2 × 10⁴ × 9.81
W = 117,720 N
Weight is acting downward
The net positive buoyant using resolving
Fb+ = Fb — W
Fb+ = 142,846.18 — 117,720
Fb+ = 25,126.18 N
The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force
W = Fb+ + Fr
W = 25,126.18 + 1100
W = 26,226.18
mg = 26,226.18
m = 26,226.18 / 9.81
m = 2673.4kg
Mass of submarine is 2673.4kg
Hello there
the answer is
Let the initial position from where the ball is kickedat angle 45 deg, be A. The Horizontal range of the ball (AC) is V2 Sin2θ / g = 38.76 Meters
thank you
Answer:
it's sqrt 175 which is pretty close to 14(sqrt 176)
Explanation:
geo mean is a*b
then square rooted
to get the mean