A diver initially moving horizontally with speed v dives dives offthe edge of a vertical cliff and lands in the water a distance
1 answer:
You might be interested in
Refer to the diagram shown below.
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
Answer:
68 readings.
Explanation:
We need to take this problem as a statistic problem where the normal distribution table help us.
We can start considerating that X is the temperature of the solution, then
For a confidence level of 90% our is 1.645
Therefore,
Substituting for and re-arrange for n, we have that n is equal to
We need to make 68 readings for have a probability of 90% and our average is within