From the freezing temperature up to about 4°C (39°F) the water CONTRACTS. That is, it becomes MORE dense. I think I read that water is the ONLY known substance whose solid phase floats in its liquid phase. That's why the cubes float in your soda and bergs float in the ocean. And if weren't so, then life on Earth would not be possible ! Oceans and lakes would freeze from the bottom up, ONE TIME, and then never thaw again.
Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
Answer:
B. 6.6%
Explanation:
The percentage error of a measurement can be calculated using the formula;
Percent error = (experimental value - accepted value / accepted value) × 100
In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.
Hence, the percentage error can be calculated thus;
Percent error = (266.5-250.0/250.0) × 100
Percent error = 16.5/250 × 100
Percent error = 0.066 × 100
Percent error = 6.6%
Answer: The correct answer is option B.
Explanation:
Mass of the sled = 10 kg
Initial speed of the sled = 2 m/s
Kinetic energy of the sled = 
Work done by the sled = 20 joules
The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.
Hence, correct answer is option B.
To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.
The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,
M,m = Counts per second
Our radios are given by
Therefore replacing we have that,
Therefore the number of counts expect at a distance of 20 cm is 19.66cps
