All categories

3 years ago

If two cars A and B are moving with velocity 60 km/hr and 80 km/hr

Physics

1 answer:

Vitek1552 [10]3 years ago

3 0

Answer:

VAB = 20km/hr

Explanation:

Given the following data;

Velocity of car A, VA = 60km/hr

Velocity of car B, VB = 80km/hr

To find the relative velocity of B w.r.t A, VAB;

Since the two cars are moving in the same direction, we have;

VAB = VB - VA

Substituting into the equation, we have;

VAB = 80 - 60

VAB = 20km/hr

Therefore, the relative velocity of car B with respect to car A is 20 kilometers per hour.

You might be interested in

Question 6 (5 points)

Ahat [919]

Answer:

its how life meets water and earth meets air.

Explanation:

4 0

3 years ago

A bar of gold has a temperature of 31°C, and a bar of aluminum has a

Katarina [22]

Different dense matters

3 0

3 years ago

Do acids or bases turn litmus paper red?

Tom [10]

Base => Blue

aciD => reD

5 0

3 years ago

Find the frequency of a wave with a period of 0.5 seconds

strojnjashka [21]

Frequency of the wave is 2 per second

Explanation:

Frequency is the number of times waves pass at a particular point of time. Here, time period = 0.5 s

Frequency is given by the formula

f = 1/T, where f is the frequency and T is the time period

⇒ f = 1/0.5 = 2 per second

8 0

3 years ago

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