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3 years ago

In which situation will two objects, each traveling with a speed of 2 m/s, have

Physics

1 answer:

nignag [31]3 years ago

4 0

Answer:

Explanation:

the relative motion will be greater when it is travelling in opposite ditection

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How long will it take a car to travel a distance of 1km if it has an average speed of 60kph

beks73 [17]

Answer:

1÷60 h

time equals distance upon speed

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3 years ago

You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same p

tatiyna

The answer is 8 kilometers per hour

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4 years ago

Read 2 more answers

Answer it asap I promise i will mark them the Brainly

nataly862011 [7]

Answer:

A) The acceleration is zero

B) The total distance is 112 m

Explanation:

Velocity vs Time Graph

It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.

The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.

A. To calculate the acceleration, we use the formula:

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

$\displaystyle a=\frac{8-8}{12-0}=0$

This confirms the previous conclusion.

B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.

Area of rectangle= base*height=12 s * 8 m/s = 96 m

Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m

The total distance is: 96 m + 16 m = 112 m

4 0

3 years ago

When a machine is used to do work, the force applied by the machine is called the effort force?

nignag [31]

Y e s i t ' s c a l l e d " t h e e f f o r t f o r c e "

H o p e t h i s h e l p s

:) L O L

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3 years ago

Two force A and B at a point at right angles. If their resultant is 50N and their sum is 70N,

kodGreya [7K]

Taking a wild guess here, but it sounds like you're asked to find |A| and |B| given the magnitude of the resultant A + B, i.e. |A + B| = 50 N, and the sum of the individual magnitudes, |A| + |B| = 70 N.

Recall that the dot product of a vector with itself is equal to the square of that vector's magnitude,

A • A = |A|²

Then

|A + B|² = (A + B) • (A + B)

|A + B|² = (A • A) + 2 (A • B) + (B • B)

|A + B|² = |A|² + 2 (A • B) + |B|²

Since A and B are perpendicular to one another, their dot product is

A • B = 0

So it follows that

|A + B|² = |A|² + |B|²

2500 N² = |A|² + |B|²

Substitute |B| = 70 N - |A| and solve for |A| :

2500 N² = |A|² + (70 N - |A|)²

2500 N² = |A|² + 4900 N² - (140 N) |A| + |A|²

2 |A|² - (140 N) |A| + 2400 N² = 0

|A|² - (70 N) |A| + 1200 N² = 0

|A|² - (70 N) |A| = - 1200 N²

|A|² - (70 N) |A| + 1225 N² = - 1200 N² + 1225 N²

(|A| - 35 N)² = 25 N²

|A| - 35 N = ± 5 N

|A| = 35 N ± 5 N

so that |A| = 30 N or |A| = 40 N. If we fix |A| to be one of these, then |B| will have the other value.

So, the magnitudes are |A| = 30 N and |B| = 40 N.

3 0

3 years ago