Answer:
0.000001 kg
Explanation:
because 1 kg equal 1,000,000 milligrams
we take
which equals 0.000001 kg
The new velocity after 4 s is 40 m/s
The height of the spaceship above the ground after 5 seconds is 1,127.5 m
The given parameters for the first question;
- initial velocity of the car, u = 76 m/s
- acceleration of the car, a = - 9 m/s²
The new velocity after 4 s is calculated as;
v = u + at
v = 76 + (-9)(4)
v = 76 - 36
v = 40 m/s
(5)
The given parameters;
- height above the ground, h = 500 m
- velocity of spaceship, u = 150 m/s
The height of the spaceship above the ground after 5 seconds is calculated as;

Learn more here: brainly.com/question/24527971
Any sample of an <em>ELEMENT</em> is made of only one type of atom.
Here are some elements:
-- Hydrogen, Helium, Neon
-- Carbon (lead in a pencil, also diamonds)
-- Oxygen, Nitrogen, Argon (All mixed together in air, but not hooked up with other atoms)
Other elements you may have heard of:
-- Silver, Gold, Copper, Iron, Lead, Aluminum, Silicon, Chlorine, Calcium, Titanium, Nickel, Tin, Platinum, Mercury, Radium, Uranium
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
Answer:
Explanation:
Given
Diameter(D)=21.5 cm
distance between plate=1.75 mm
and we know capacitance is

and we know charge is
Q=CV
and charge density (![\sigma [tex])[tex]=\frac{Q}{A}=\frac{\epsilon _0AV}{Ad}=\frac{\epsilon V}{d}](https://tex.z-dn.net/?f=%5Csigma%20%5Btex%5D%29%5Btex%5D%3D%5Cfrac%7BQ%7D%7BA%7D%3D%5Cfrac%7B%5Cepsilon%20_0AV%7D%7BAd%7D%3D%5Cfrac%7B%5Cepsilon%20V%7D%7Bd%7D)
Electric field(E)
Energy density
Energy desity
Energy desity![=\frac{1}{8\pi \cdot 9\times 10^9}\times \left [ \frac{12\times 10^3}{1.75}\right ]^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B8%5Cpi%20%5Ccdot%209%5Ctimes%2010%5E9%7D%5Ctimes%20%5Cleft%20%5B%20%5Cfrac%7B12%5Ctimes%2010%5E3%7D%7B1.75%7D%5Cright%20%5D%5E2)
