Answer:
14.625 years
Explanation:
Because acceleration/deceleration are negligent we can simplify this problem into 2 basic parts:
First: how long before the astronaut arrives? Since the speed of the craft is 0.8 times the speed of light and the distance to travel is 6.5 light-years, the equation to answer this part is:
0.8t=6.5
solve for t by dividing by 0.8
t=6.5/0.8 or 8.125
so, she will arrive in 8.125 years.
Then, once she sends her message, it will travel towards earth at 1 times the speed of light or:
1.0t=6.5 which is just t=6.5
so, her message will arrive back to Earth (8.125+6.5) 14.625 years after takeoff.
Because the two paths are perpendicular, therefore the
target proton's new path must be at 30 degrees from the original
direction.
Using the law of conservation of momentum about the original direction:
m (400 m/s) = m (v1) cos(60) + m (v2) cos(30)
Cancelling m since the two protons have similar mass.
(v1)cos(60) + (v2)cos(30) = 500 m/s ---> 1
Now by using the law conservation of momentum perpendicular to the original
direction:
m (0 m/s) = m (v1) sin(60) – m (v2) sin(30)
Which simplifies to:
(v1)sin(60) - (v2)sin(30) = 0 m/s
v2 = v1 * sin(60) / sin(30) = v1 * sqrt(3) ---> 2
Plugging equation 2 to equation 1:
(v1) (1/2) + (v1 * sqrt(3)) sqrt(3)/2 = 500 m/s
(1/2) (v1) + (3/2) (v1) = 500 m/s
2 (v1) = 500 m/s
v1 = 250 m/s
Thus, from equation 2:
v2 = v1*sqrt(3) = (250 m/s) sqrt(3) = 433.01 m/s
So,
A. The target proton's speed is about 433 m/s
B. The projectile proton's speed is 250 m/s
<span>tidal pattern
</span>
<span>tidal force
</span>
Answer:
Look Below -->
Explanation:
a. She traveled 10 km, add 4.5 km + 5.5 km = 10 km (Distance is the total units travelled, so just add them all up :) )
b. Her displacement is 0 km because she went back home. (Displacement is the difference between the end and starting points)
c. 3 km/hr (30 minutes / 10 km)