Answer:
The one with equal forces on both sides, D is your answer
Explanation:
A, 5 and 10 No
B 13 and 8 no
C 5 and 8 no
D 6 and 6 YES
Answer:
The wavelength of the waves created in the swimming pool is 0.4 m
Explanation:
Given;
frequency of the wave, f = 2 Hz
velocity of the wave, v = 0.8 m/s
The wavelength of the wave is given by;
λ = v / f
where;
λ is the wavelength
f is the frequency
v is the wavelength
λ = 0.8 / 2
λ = 0.4 m
Therefore, the wavelength of the waves created in the swimming pool is 0.4 m
Answer:
4 days
either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2
n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2
Explanation:
Answer:
the terminal velocity of 14 nested coffee filters is 3.2 m/s
Explanation:
Given the data in the question;
we know that;
The terminal velocity is proportional to the square root of weight.
v ∝ √W
v = k√W
the proportionality constant depends upon the surface area and the density of the medium (like air). The coffee filters can be stacked such that the resulting area is roughly unchanged. So, the constant of proportionality k is also unchanged
v/√W = constant
v₂/√W₂ = v₁/√W₁
v₂ = v₁√(W₂ / W₁ )
given that;
v₁ = 0.856 m/s,
W₂ = 14W₁; meaning 14 coffee filters have 14 times the weight of a single coffee filter
so we substitute
v₂ = 0.856 √(14W₁ / W₁ )
v₂ = 0.856 √( 14( W₁/W₁)
v₂ = 0.856 √( 14(1)
v₂ = 0.856 √( 14 )
v₂ = 0.856 × 3.741657
v₂ = 3.2 m/s
Therefore, the terminal velocity of 14 nested coffee filters is 3.2 m/s
Answer:
a)
a = 2 [m/s^2]
b)
a = 1.6 [m/s^2]
c)
xt = 2100 [m]
Explanation:
In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.
a)
When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

where:
Vf = final velocity = 40 [m/s]
Vi = initial velocity = 0 (starting from rest)
a = acceleration [m/s^2]
t = time = 20 [s]
40 = 0 + (a*20)
a = 2 [m/s^2]
The distance can be calculates as follows:

where:
x1 = distance [m]
40^2 = 0 + (2*2*x1)
x1 = 400 [m]
Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.
v = x2/t2
where:
x2 = distance [m]
t2 = 30 [s]
x2 = 40*30
x2 = 1200 [m]
b)
Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.
0 = 40 - (a *25)
a = 40/25
a = 1.6 [m/s^2]
The distance can be calculates as follows:

0 = (40^2) - (2*1.6*x3)
x3 = 500 [m]
c)
Now we sum all the distances calculated:
xt = x1 + x2 + x3
xt = 400 + 1200 + 500
xt = 2100 [m]