Answer:
3500N
Explanation:
Given parameters:
Mass of driver = 50kg
Speed = 35m/s
Time = 0.5s
Unknown:
Average force the seat belt exerts on her = ?
Solution:
The average force the seat belt exerts on her can be deduced from Newton's second law of motion.
F = mass x acceleration
So;
F = mass x 
F = 50 x 
= 3500N
The equation 
(option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:
Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:
When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:
Therefore, the equation represents the horizontal momentum (option 3).
Learn more about linear momentum here:
I hope it helps you!
2e min :)) pls park braliest
Answer:
1.05 J.
Explanation:
Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as
Ek = 1/2mv²................. Equation 1
Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.
But,
v = αr .......................... Equation 2
Where α = angular velocity of the rod, r = radius of the circle.
Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.
Substitute into equation 2
v = 3.6(0.6)
v = 2.16 m/s.
Also given: m = 450 g = 0.45 kg.
Substitute into equation 1
Ek = 1/2(0.45)(2.16²)
Ek = 1.05 J.
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
