Answer:
The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.
Explanation:
By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:
Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.
When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:
By replacing the known values it is possible to find the plate's weight:
When the plate kept to 30° from the vertical the moment equation balance is written as:
The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:
Answer:
The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change
Explanation:
Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.
Answer:
a = 5.05 x 10¹⁴ m/s²
Explanation:
Consider the motion along the horizontal direction
= velocity along the horizontal direction = 3.0 x 10⁶ m/s
t = time of travel
X = horizontal distance traveled = 11 cm = 0.11 m
Time of travel can be given as
inserting the values
t = 0.11/(3.0 x 10⁶)
t = 3.67 x 10⁻⁸ sec
Consider the motion along the vertical direction
Y = vertical distance traveled = 34 cm = 0.34 m
a = acceleration = ?
t = time of travel = 3.67 x 10⁻⁸ sec
= initial velocity along the vertical direction = 0 m/s
Using the kinematics equation
Y = t + (0.5) a t²
0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²
a = 5.05 x 10¹⁴ m/s²
Answer:
The mass of the cube is 420.8 kg.
Explanation:
Given that,
Length of edge = 38.9 cm
Density 
We need to calculate the volume of cube
Using formula of volume
We need to calculate the mass of the cube
Using formula of density
Hence, The mass of the cube is 420.8 kg.
<span>With a half-life of 700 million years, U-235 would have had twice as much mass at a time 700 MYA. This would have put the mass at 100kg at that time. Going back another 50 million years would be (50/700) or 1/14 of the half-life, or (1/2 * 1/14), or 1/28 of the total mass. 1/28 of 100kg is 3.57kg, so the amount present at the 750MYA mark would be approximately 103.57kg of U-235.</span>
