Answer:
r = 0.86
Explanation:
Correlation coefficients are the strength of the relationship between two variables.
Correlations can indicate anywhere between
- 1 - for a strong positive relationship.
- -1 - for a strong negative relationship.
- 0 - for no relationship at all.
Looking at sample correlation coefficient formula which says
= 
÷ (
× 
)
where and are the sample deviations and is the sample covariance, all of which will remain the same for Maria and John.
Hence, John's correlation will be approximately 0.86 since he would have approximately the same measurement as Maria's measurement when Maria's measurement is converted from centimeters to inches.
DaddyFed is right, it would be all of them.
Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
Answer:
15.67 m/s
Explanation:
The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.
Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by
where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:
Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
