The best and most correct answer among the choices provided by the question is the second choice. The rope will not be broken until the net force acting on it is not equal to zero anymore. I hope my answer has come to your help. God bless and have a nice day ahead!

5 0

4 years ago

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2. A student drives 7.8-km trip to school and averages a speed of

Alekssandra [29.7K]

Answer:

The total time is: t=451.22 sec

The average speed is: V=34.57 m/s

Explanation:

Average speed

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

$\displaystyle V=\frac{x}{t}$

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

$\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec$

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

$\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec$

The total time is:

$t=239.26\ sec+211.96\ sec=451.22\ sec$

$\boxed{t=451.22\ sec}$

The average speed is:

$\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s$

$\boxed{\displaystyle V=34.57\ m/s}$

6 0

3 years ago

GO TO MY LATEST QUESTION AND ANSWER I WILL BRAINLIST AND GIVE 35 POINTS TO THE CORRECT ANSWER

REY [17]

y e s s s s i r r r r r r

8 0

3 years ago

A car moving at some speed hits the brakes and skids to a stop after 13 m on a level road. If the coefficient of friction for th

vesna_86 [32]

Answer:

12.974 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction =0.66

a = Acceleration = $\mu g$

$v^2-u^2=2as\\\Rightarrow -u^2=2\mu gs-v^2\\\Rightarrow u=\sqrt{v^2-2\mu gs}\\\Rightarrow u=\sqrt{0^2-2\times -(9.81\times 0.66)\times 13}\\\Rightarrow u=12.974\ m/s$

Car's original speed before braking was 12.974 m/s

6 0

3 years ago