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:
brainly.com/question/17228388
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The acceleration of the wagon along the ground is 3.6 m/s².
To solve the problem above, we need to use the formula of acceleration as related to force and mass.
Acceleration: This can be defined as the rate of change of velocity.
⇒ Formula:
- Fcos∅ = ma................. Equation 1
⇒ Where:
- F = Force
- ∅ = angle above the horizontal
- m = mass of the wagon
- a = acceleration of the wagon
⇒ make a the subject of equation 1
- a = Fcos∅/m..................... Equation 2
From the question,
⇒ Given:
⇒ Substitute these values into equation 2
- a = 44(cos35°)/10
- a = 44(0.8191)/10
- a = 3.6 m/s²
Hence, The acceleration of the wagon along the ground is 3.6 m/s²
Learn more about acceleration here: brainly.com/question/9408577
Answer:
Therefore, the revolutions that each tire makes is:
![\Delta \theta=22\: rev](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D22%5C%3A%20rev)
Explanation:
We can use the following equation:
(1)
The angular acceleration is:
![a_{tan}=\alpha R](https://tex.z-dn.net/?f=a_%7Btan%7D%3D%5Calpha%20R)
![\alpha=\frac{1.9}{0.325}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B1.9%7D%7B0.325%7D)
![\alpha=5.85\: rad/s^{2}](https://tex.z-dn.net/?f=%5Calpha%3D5.85%5C%3A%20rad%2Fs%5E%7B2%7D)
and the initial angular velocity is:
![\omega_{i}=\frac{v}{R}](https://tex.z-dn.net/?f=%5Comega_%7Bi%7D%3D%5Cfrac%7Bv%7D%7BR%7D)
![\omega_{i}=\frac{27.2}{0.325}](https://tex.z-dn.net/?f=%5Comega_%7Bi%7D%3D%5Cfrac%7B27.2%7D%7B0.325%7D)
![\omega_{i}=83.69\: rad/s](https://tex.z-dn.net/?f=%5Comega_%7Bi%7D%3D83.69%5C%3A%20rad%2Fs)
Now, using equation (1) we can find the revolutions of the tire.
![\Delta \theta=135.47\: rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D135.47%5C%3A%20rad)
Therefore, the revolutions that each tire makes is:
![\Delta \theta=22\: rev](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D22%5C%3A%20rev)
I hope it helps you!
Answer:
a₁ = 0.63 m/s² (East)
a₂ = -1.18 m/s² (West)
Explanation:
m₁ = 95 Kg
m₂ = 51 Kg
F = 60 N
a₁ = ?
a₂ = ?
To get the acceleration (magnitude and direction) of the man we apply
∑Fx = m*a (⇒)
F = m₁*a₁ ⇒ 60 N = 95 Kg*a₁
⇒ a₁ = (60N / 95Kg) = 0.63 m/s² (⇒) East
To get the acceleration (magnitude and direction) of the woman we apply
∑Fx = m*a (⇒)
F = -m₂*a₂ ⇒ 60 N = -51 Kg*a₂
⇒ a₂ = (60N / 51Kg) = -1.18 m/s² (West)
For every case we apply Newton’s 3
d Law
Define displacement and I'll help you