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)
Alright, so solving for the net force is a rather simple and easy step. I want to briefly explain how you determine the net force, but before I do, let's examine the problem together. What has been provided to us? What direction is it going? Well, we can look at the information!
Amanda's Force: 5 N
Jeremiah's Force: 10 N
They're both pushing on the books, but in different directions. Left and right. If Amanda is pushing the books to the right 5 newtons, and Jeremiah is pushing the books to the left with 10 newtons, that means the net force is 5. The book is being pushed to the left! We can say that because Jeremiah's force is much larger than his sisters.
We can determine the net force from the following;
Net Force → 10 N - 5 N = 5 N
In conclusion, your answer should be a total of five (5) newtons
<u>Answer</u>
4.8212×10⁻¹¹ N
<u>Explanation</u>
The gravitational force F, between 2 masses m₁ and m₂ is given by:
F = (Gm₁m₂)/d²
Where G = 6.673 x 10⁻¹¹ N m²/kg² and d is the distance between the 2 masses.
F = (Gm₁m₂)/d²
= (6.673 x 10⁻¹¹ × 0.425 × 0.425)/0.500²
= 1.2053×10⁻¹¹/0.25
= 4.8212×10⁻¹¹ N
Answer:
12m/s
Explanation:
x = (21 + 22t – 6 t2) m,
We can express change bin velocity as
dx/dt= -12t + 22
But Velocity is changing
interval t = 1 s
t = 4 s
at t = 1, x = 21 + 22(1) -6(1)^2
=37
at t = 4, x = 21 + 22(4) -6(4)^2
=13
Distance travelled at interval t = 1 s
t = 4 s
X2 - X1=13- 37. = - 24m
Velocity= displacement/ time
= 24/(3-1)=12 m/s
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
