Answer:
answer = 12.87 km/h
Step-by-step explanation:
Given
Ship A is sailing east at 25 km/h = 
ship B is sailing north at 20 km/h =
here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively
so we get x = 4 ×25 =100 km/h
y = 4× 20 = 80 km/h
let z is the distance between the ships, we need to find 
at t = 4 hr
At noon, ship A is 130 km west of ship B (12:00 pm)
so equation will be
derivative first equation w . r. to t we get
![\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]](https://tex.z-dn.net/?f=%5Cfrac%7Bdz%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7Bz%7D%5B%28x%20-130%29%5Cfrac%7Bdx%7D%7Bdt%7D%20%2By%5Cfrac%7Bdy%7D%7Bdt%7D%5D)
Put them all in to decimals so dived 1 by 5 and 32 by 100 then they will all be decimals and you can more essay compare them
Answer:
The net forces exerted on the horse and cart are not the same, so they are not balanced forces.
Step-by-step explanation:
Please see the Newton's 2nd Law which states that an object accelerates if there is a net or unbalanced force on it. In this scenario there is just one force exerted on the wagon i.e: the force that the horse exerts on it. The wagon accelerates because the horse pulls on it. And the amount of acceleration equals the net force on the wagon divided by its mass.
As there are two forces the push and pull the horse; the wagon pulls the horse backwards, and the ground pushes the horse forward. The net force is determined by the relative sizes of these two forces.
If the ground pushes harder on the horse than the wagon pulls, there is a net force in the forward direction, and the horse accelerates forward, and if the wagon pulls harder on the horse than the ground pushes, there is a net force in the backward direction, and the horse accelerates backward.
If the force that the wagon exerts on the horse is the same size as the force that the ground exerts, the net force on the horse is zero, and the horse does not accelerate.
Answer:
The value of a is 101°, b is 79°, c is 83° and d is 97°
Acceleration is the change in velocity with respect to the time. The acceleration of the car at segment C is -30 meter per second squared. Hence the option B is the correct option.
<h3>
Given information-</h3>
Segment A runs from 0 seconds 0 meters per second to 1 seconds 30 meters per second.
Segment B runs to 3 seconds 30 meters per second.
Segment C runs to 6 seconds 10 meters per second.
Segment D runs to 7 seconds 10 meters per second.
Segment E runs to 10 seconds 20 meters per second.
<h3>Acceleration</h3>
Acceleration is the change in velocity with respect to the time. Acceleration of a vector quantity which means it has both magnitude and the direction. It can be given as,
Here 
denotes the time and 
denotes the velocity of the body.
Acceleration at segment C,
Thus the acceleration of the car at segment C is -30 meter per second squared. Hence the option B is the correct option.
Learn more about the acceleration here;
brainly.com/question/2437624
