Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
The resistance expected of the heater is 50.1 ohms.
<h3>What is resistance?</h3>
Resistance can be defined as the opposition to the flow of electric current in an electric circuit. The S.I unit of resistance is Ohms (Ω).
To calculate the resistance of the heater, we use the formula below.
<h3>Formula:</h3>
- R = V²/P............. Equation 1
Where:
- R = Resistance of the heater
- P = Power of the heater
- V = Voltage supplied to the heater
From the question,
Given:
- V = 480 V
- P = 4.6 kW = 4600 W
Substitute these values into equation 1
- R = (480²)/4600
- R = 50.1 ohms.
Hence, the resistance expected of the heater is 50.1 ohms.
Learn more about resistance here: brainly.com/question/17563681
8 ∙ 10^-4 / 2 ∙ 10^2 = (8/2) ∙ ((10^-4)*(10^-2)) = <span>4 ∙ 10^-6</span>
D. frequency of the corresponding light wave
The first successful flyby of Venus was performed by NASA's Mariner 2 spacecraft on 14 December 1962, following failed attempts by both the Soviet Union and the USA. The first successful landing was the Soviet Venera 4 lander, which touched down on the surface on 18 October 1967
