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
Answer:
FG and FP
Explanation:
Gravitational Force(FG) because the box is being pulled down and a resistance force pushing up because he is pushing(FP) up on the box.
I hope this helps!
Answer:
The magnetic field strength inside the solenoid is
.
Explanation:
Given that,
Radius = 2.0 mm
Length = 5.0 cm
Current = 2.0 A
Number of turns = 100
(a). We need to calculate the magnetic field strength inside the solenoid
Using formula of the magnetic field strength
Using Ampere's Law

Where, N = Number of turns
I = current
l = length
Put the value into the formula


(b). We draw the diagram
Hence, The magnetic field strength inside the solenoid is
.
Answer:
kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Explanation:
Given the data in the question;
we know that;
Kinetic energy = 1/2.mv²
given that mass of the object is doubled; m1 = 2m
speed is halved; v1 = V/2
Now, New kinetic energy will be; 1/2.m1v1²
we substitute
Kinetic Energy = 1/2 × 2m × (v/2)²
Kinetic Energy = 1/2 × 2m × (v²/4)
Kinetic Energy = 1/2 × m × (v²/2)
Kinetic Energy = 1/2 [ 1/2mv² ]
Kinetic Energy = 1/2 [ KE ]
Therefore; kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer