Voltage = (current) x (resistance) (Ohm's law)
Voltage = (15 A) x (18 ohms)
<em>Voltage = 270 volts</em>
the answer is D. A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion → Each fluorine atom becomes a -1 ion
because magnesium only has 2 valence electrons, and it would be easier to lose electrons rather gain.
i just took the test, its right.
Q: A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula:
h(t)=100+45t−4.9t2
(a)
What is its height after 3 seconds?
(b)What is its velocity after 3 seconds?
Answer:
(a) 190.9 m.
(b) 15.6 m/s upward
Explanation:
Given:
h(t) = 100 + 45t - 4.9t²
The height after 3 seconds,
t = 3 s
Substitute the value of t in to the equation above.
h(3) = 100+45(3)-4.9(3)²
h(3) = 100+135-44.1
h(3) = 190.9 m
Therefore the height after 3 seconds = 190.9 m.
(b) Velocity after 3 seconds
The velocity is obtained by differentiating h(t) with respect to time
v = dh(t)/dt
dh(t)/dt = 45-9.8t
v = 45 - 9.8t ......................................... Equation 1
t = 3 s.
Substitute the value of t into the equation above,
v = 45 - 9.8(3)
v = 45- 29.4
v = 15.6 m/s
Thus the velocity after 3 seconds = 15.6 m/s upward
Refract mean to break. the quarter would look broken. hope this helps
Answer:
a) x = 0.0175 m and b) v = 1.178 m/s
Explanation:
a) Let's analyze the exercise gives us a compressed spring, which would have elastic energy and a glider that is initially at rest and then released acquiring kinetic and potential energy, as there is no friction, the energy is conserved
Initial
Emo = Ue = ½ k x²
Final
Emf = K + U = ½ m v² + mg y
Emo = Emf
½ k x² = ½ m v² + mg y
At the highest point the speed is zero (v = 0). Let's calculate height trigonometry
sin 44 = y / L
y = L sin44
½ k x² = 0 + m g y
x² = 2m g L sin44 / k
x = √ (2 8.00 10⁻² 1.6 sin44 / 580)
x = 0.0175 m
b) When the glider is 0.60m it is no longer in contact with the spring that contracted only 0.0175m
Let's write the mechanical energy at this point
Em2 = K + U
Em2 = ½ m v² + mg y₂
L2 = 0.60 m
y₂ = L2 sin44
Emo = Em2
½ k x² = ½ mv² + mgh
v² = (½ k x² -mgh) 2 / m
v = √ [(½ k x² - mg L₂ sin44) 2 / m]
Let's calculate
v = √ [(½ 580 0.0175² - 8 10⁻² 0.6 sin44) 2/8 10⁻²]
v = √[ (0.08881 – 0.0333) 25]
v = 1.178 m/s