Answer:
2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]
Explanation:
2a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2032%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%201%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D32%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A1%5Bh%5D%5C%5Cx%3D32%5Bmil%7D)
2b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B420%7D%7B840%7D%5C%5C%20t%3D0.5%5Bh%5D)
3a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B35%7D%7B14%7D%5C%5C%20t%3D2.5%5Bh%5D)
3b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2074%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%202.5%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D74%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A2.5%5Bh%5D%5C%5Cx%3D185%5Bmil%7D)
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:
F = 0.1575 N
Explanation:
When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.
In this moment then
Sphere one has a charge = Q/2
Sphere three has a charge = Q/2
Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.
How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q
Sphere two has a charge = 3/4Q
Sphere three has a charge = 3/4Q
The electrostatic force that acts on sphere 2 due to sphere 1 is:
F = 
F= 
how 
= 0.42
Then
F = 
F = 0.1575 N
The final velocity is +15.0 m/s
Explanation:
The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:
where
v is the velocity at time t
u is the initial velocity
a is the acceleration
t is the time
For the cart in this problem, we have:
u = +3.0 m/s (initial velocity)
(acceleration)
t = 8.0 s (time)
Substituting, we find the final velocity:
Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
#LearnwithBrainly
