Answer:
58.9 N
Explanation:
The wood is buoyed up by the mass of the water is displaces
wood volume = .030 m^3 = 30 000 cm^3
mass of water displaced = 30 000 g = 30 kg
wood mass = 800 kg/m^3 * .030 m^3 = 24 kg
so tension in string = mg = ( 30 kg - 24 kg) * 9.81 m/s^2 = 58.9 N
10 cm3 because the density equition is d = m/v
Answer:
e) True. Measure the own values, so everything seems normal
Explanation:
In the case of special relativity, it is explicitly stated that the speed of light is constant and equal to c for all inertial observers. For this reason the measures of time and length are no longer the same for observers moving with respect to each other.
We call the time and the proper length the magnitude measured for an observer who does not move with respect to the measurement system.
In this case the astronomer is on the ship, for him he does not feel the movement of it, they are at rest with respect to each other. Therefore, their measurements are the so-called ones, this means that their values do not change since the two go at the same speed.
In examining the final statements we have
A) False. The mass measures not
B) False. Measure own length
C) False
D) false
e) True. Measure the own values, so everything seems normal
Answer:
3.4093
Explanation:
NPSHa = hatm + hel + hf +hva
the elevation head is the hel
friction loss head is hf
NPSHa is the head of vapour pressure of fluid
atmospheric pressure head is hatm
log₁₀P* = 
log₁₀Pv = 
= 4.07827 - 1343.943/333.377
=4.07827 - 4.0313009
= 0.0469691
we take the log
p* = 1.114218
we convert this value to get 111421.8
hvap = 111421.8 * 1/776.14 * 1/9.81
= 14.63
hatm = 1.1 *101325/1 * 1/9.81 *1/776.14
=14.64
hf = 7000/1 * 1/776.14 * 1/9.81
= 0.9193
NPSHa = 2.5
hel = 0.9193 + 2.5 + 14.63 - 14.64
hel = 3.4093
<u>Answer:</u>
Speed at which the ball leaves the launcher = 6.34 m/s
<u>Explanation:</u>
The readings of height above the floor reached by the ball = 2.32, 2.26 and 2.37
Mean value 
Initial height of launcher = 27 cm = 0.27 cm
So maximum height of projection = 2.317-0.27 = 2.047 meter
We have equation of motion, 
, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
In this case we need to consider only vertical factors.
Vertical displacement = 2.047 m, acceleration = acceleration due to gravity = 
, final velocity = 0 m/s, we need to calculate initial velocity.
So, speed at which the ball leaves the launcher = 6.34 m/s
