The velocities and the speed build a triangle, where the 1.7 m/s are the hypotenuse and the x-velocity and y-velocity are the other sides.
<span>So the x-velocity is: speed*cos(angle) </span>
<span>now plug in </span>
<span>x=1.7 m/s * cos(18.5)=1.597 m/s </span>
Measuring density: Measure the mass (in grams) of each mineral sample available to you. The mass of each sample is measured using a balance or electronic scale. Record mass on a chart.
If Calcium lost two electrons, it would have the same number of electrons as Argon which has 18 electrons.
Answer:
A. Vx = 3.63 m/s
B. Vy = -45.73 m/s
C. |V| = 45.87 m/s
D. θ = -85.46°
Explanation:
Given that position, r, is given as:
r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk
Velocity is the derivative of position, r:
V = dr/dt = 3.63 - 11.46t^j
A. x component of velocity, Vx = 3.63 m/s
B. y component of velocity, Vy = -11.46t
t = 3.99 secs,
Vy = - 11.46 * 3.99 = -45.73 m/s
C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]
|V| = √(2091.2329 + 13.1769)
|V| = √(2104.4098)
|V| = 45.87 m/s
D. Angle of the velocity relative to the x axis, θ is given as:
tanθ = Vy/Vx
tanθ = -45.73/3.63
tanθ = -12.6
θ = -85.46°
Answer:
Increases
Explanation:
Since power P=IV
Then it means when current increases, the power increases hence brightness increases. I represent current, P is power and v is voltage.
Current of capacitor when in series connection is given by
where I is current across capacitor, f is frequency, C is capacitance and v is voltage across capacitance. From this second formula, it is evident that an increase in capacitance increases the current across the capacitor. Therefore, if current increases, power also increases leading to an increase in brightness
