At 8:00 pm, the velocity of the storm is 55 mi northeast. Assuming that the direction is exactly northeast, the angle is 45°
At 11:00 pm, the velocity is 75 mi north. The angle is 90°
In vector form
55 ∠ 45°
and
75 ∠ 90°
The magnitude and direction of the average velocity is
(55 ∠ 45° + 75 ∠ 90° ) / 3
Answer:
1.3m/s
Explanation:
Data given,
Mass,m=1.0kg,
Amplitude,A=0.10m,
Frequency,f=2.0Hz.
From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

we can express the velocity by the derivative of the displacement,
Hence

at equilibrium, the velocity becomes

Hence if we substitute values we arrive at

Answer:
the answer is A
Explanation: The vertical temperature line indicates no change in temperature at those altitudes.
Answer:
The acceleration of the ball is 4.18 [m/s^2]
Explanation:
By Newton's second law we can find the acceleration of the ball
![F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5Cwhere%3A%5C%5CF%20%3D%20force%20applied%20%5BN%5D%20or%20%5Bkg%2Am%2Fs%5E2%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Ca%20%3D%20acceleration%20%5Bm%2Fs%5Es%5D)
Now we have:
![a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]](https://tex.z-dn.net/?f=a%20%3D%20F%2Fm%5C%5Ca%20%3D%20%5Cfrac%7B1.8%20%5Bkg%2Am%2Fs%5Es%5D%7D%7B0.43%5Bkg%5D%7D%20%5C%5Ca%20%3D%204.18%20%5Bkg%5D)
Answer:
B. 450 feet
Explanation:
Due to the angle at which high beam headlights illuminate, they can illuminate the road for about 450 feet.