Answer:
C. 13.89 meters/second
Explanation:
If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:
Vv=80*sin(10)=80*0.1736=13.888 m/s.
So the vertical component of the velocity of the train is Vv=13.888 m/s.
Answer:
The area under a velocity-time graph is the displacement. Velocity can be negative if an object is moving backwards. The displacement can also be negative. An area beneath the x-axis has a negative value.
Answer:
Q = 2687130 J
Explanation:
m = mass of block of ice = 5 kg
= Initial temperature of ice block = - 27 °C
= final temperature of water = 35 °C
= specific heat of ice = 2108 J/(Kg °C)
= Latent heat of fusion of ice = 334000 J/kg
= specific heat of water = 4186 J/(Kg °C)
Heat added is given as
Though it would be very cold, you could put the human manned station of one of Uranus' moons.
Answer:
1.424 μC
Explanation:
I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),
tension T (acting along the string - to the pivot point), and
F (electric force – acting along the line connecting the charges).
We then have something like this
x: T•sin α = F,
y: T•cosα = mg.
Dividing the first one by the second one we have
T•sin α/ T•cosα = F/mg, ultimately,
tan α = F/mg.
Since we already know that
q1=q2=q, and
r=2•L•sinα,
k=9•10^9 N•m²/C²
Remember,
F =k•q1•q2/r², if we substitute for r, we have
F = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • √(m•g•tanα/k)=
=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =
q = 0.486 • √(8.61•10^-12)
q = 0.486 • 2.93•10^-6
q = 1.424•10^-6 C
q = 1.424 μC.