Explanation:
Both oceanic crust and continental crust are less dense than the mantle, but oceanic crust is denser than continental crust. ... Because continental crust is less dense than oceanic crust it floats higher on the mantle, just like a piece of Styrofoam floats higher on water than a piece of wood does.
The continental crust is by far the older of the two types of crust.
Hope this helps you out ; )
Answer:
1 m/s²
Explanation:
From the question,
Using
a = (v-u)/t.................... Equation 1
Where a = accelartion of the bicycle, v = Final velocity, u = initial velocity, t = time.
Given: v = 15 m/s, u = 5 m/s, t = 10 s
Substitute these values into equation 1
a = (15-5)/10
a = 10/10
a = 1 m/s²
Hence the acceleration of the bicycle is 1 m/s²
Answer:
Explanation:
To calculate an effect size, called Cohen's d , for the one-sample t-test you need to divide the mean difference by the standard deviation of the difference, as shown below. Note that, here: sd(x-mu) = sd(x) . μ is the theoretical mean against which the mean of our sample is compared (default value is mu = 0).
(a) 
The gravitational potential energy of the two-sphere system is given by
(1)
where
G is the gravitational constant
is the mass of sphere A
is the mass of sphere B
r = 1.8 m is the distance between the two spheres
Substitutign data in the formula, we find
and the sign is negative since gravity is an attractive force.
(b) 
According to the law of conservation of energy, the kinetic energy gained by sphere B will be equal to the change in gravitational potential energy of the system:
(2)
where
is the initial potential energy
The final potential energy can be found by substituting
r = 1.80 m -0.60 m=1.20 m
inside the equation (1):
U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.2 m}=-5.22\cdot 10^{-7} J
So now we can use eq.(2) to find the kinetic energy of sphere B:
Answer:
x=2.4t+4.9t^2
Explanation:
This equation is one of the kinematic equations to solve for distance. The original equation is as follows:
X=Xo+Vt+1/2at^2
We know that the ball starts at rest meaning that its initial velocity and position is zero.
X=0+Vt+1/2at^2
Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.
X=Vt+4.9t^2
Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.
Now, substitute in your velocity value.
X=2.4t+4.9t^2
