Speed of the Enemy sub .....> <span> 25.0 km/hr which is (25/3.6) m/s = 6.94 m/s
We use the expression Speed = Distance/ time to find the distance
D = Speed*Time = 6.94 m/s * 3.5 = 24.305 m
The enemy sub will be 24.305 m closer </span>
The electron is a type of low-mass, very negatively charged with a particle. As such, it can easily be deflected by passing close to other electrons or the positive nucleus of an atom. m = mass of an electron in kg = 9.10938356 × 10-31 kilograms. e = magnitude of the charge of an electron in coulombs = 1.602 x 10-19 coulombs. Hope this helps!
Answer:
9.965 nF
Explanation:
The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m
So, C = εA/d
C = ε2πrL/d
Substituting the of the values variables into the equation, we have
C = ε2πrL/d
C = 24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m
C = 9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m
C = 996463 × 10⁻¹⁴ F
C = 9.96463 × 10⁻⁹ F
C = 9.96463 nF
C ≅ 9.965 nF
Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
Answer: The drag force goes up by a factor of 4
Explanation:
The <u>Drag Force</u> equation is:
(1)
Where:
is the Drag Force
is the Drag coefficient, which depends on the material
is the density of the fluid where the bicycle is moving (<u>air in this case)
</u>
is the transversal area of the body or object
the bicycle's velocity
Now, if we assume
,
and
do not change, we can rewrite (1) as:
(2)
Where
groups all these coefficients.
So, if we have a new velocity
, which is the double of the former velocity:
(3)
Equation (2) is written as:
(4)
Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>