The correct value for the rocket speed measured by an observer on earth would be = 14c
<h3>Calculation and of the rocket speed</h3>
To calculate the speed of the rocket we should observe that the rocket and the missile are in a relative motion which are moving on the same direction.
The speed of the rocket = 12c
The speed of the missile = 12c
The individual speed of earth =0c
But, the relative speed of the missile and the Earth is the sum of individual speeds.
Thus, the speed of rocket as measured by an observer on the Earth would be 0 + 14c = 14c.
Therefore, the correct value for the rocket speed measured by an observer on earth would be =14c
Learn more about relative motion here:
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Answer:
There is no friction between the card and the cup.
Explanation:
Answer:
In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.
Explanation:
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Answer:
10.15Ω
Explanation:
From ohm's law,
V = IR...................... Equation 1
Where V = Voltage, I = current, R = resistance.
Assume the voltage across the resistance = V,
Given: I = 6.3 A
Substitute into equation 1
V = 6.3R.................. Equation 2
When an additional resistance of 3.4 Ω is inserted in series with R,
The voltage remain the same, but the current changes
Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A
Also from ohm' law,
V = I'Rt............... Equation 3
Substitute the value of I' and Rt into equation 3
V = 4.72(R+3.4)............... Equation 5.
Divide equation 2 by equation 5
V/V = 6.3R/4.72(R+3.4)
1 = 1.335R/(R+3.4)
1 = 1.335R/(R+3.4)
R+3.4 = 1.335R
3.4 = 1.335R-R
3.4 = 0.335R
R = 3.4/0.335
R = 10.15Ω
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
