As these are distances created by moving in a straight line, using a trigonometric analysis can solve the missing single straight-line displacement. Looking at the 48m and 12m movements as legs of a triangle, obtaining the hypotenuse using the pythagorean theorem will yield us the correct answer.
This is shown below:
c^2 = 48^2 + 12^2
c = sqrt(2304 + 144)
c = sqrt(2448)
c = 49.48 m
To obtain the angle at which Anthony walks 49.48, we obtain the arc tangent of (12/48). This is shown below:
arc tan (12/48) =14.04 degrees.
Therefore, Anthony could have walked 49.48 m towards the S 14.04 W direction.
we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
Answer:
The newton (lowercase n!) is a derived unit because its definition consists of multiplication of three defined base units and nothing else. Its meaning is thus derived, not independently defined.
In physics, a force is said to do work<span> if, when acting, there is a displacement of the point of application in the direction of the force. It is calculated by the formula W = f x d. Therefore, we calculate the problem above as follows:
W = f x d
4000 = f x 20
f = 200 N <----- FIRST OPTION</span>
Answer:
It requires <u>1.9 seconds</u> to reach maximum height.
Explanation:
As per given question,
Initial velocity (U) =19 m/s
Final velocity (V) = 0 m/s

Maximum height = S
Time taken is "t"
<u>Calculating time taken to reach maximum height:</u>
We know that time taken to reach the maximum height is calculated by using the formula V = U + at
Substitute the given values in the above equation.
Final velocity is “0” as there is no velocity at the maximum height.



t = 1.9 seconds.
The time taken to reach maximum height is <u>1.9</u> seconds.
<u>Calculating maximum height</u>:

Solving the equation we will get the value of S

-361 = -20S
Negative sign cancel both the sides.

S = 18.05 m
Maximum height is 18.05 m .