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3 years ago

If an object moves 40 m north, 40 m west, 40 m south, and 40 m east, what's the total displacement?

2 answers:

8 0

The displacement of the object is 0, because the object ends up in back in its starting position. The 40 m north and 40 m south will cancel out. Similarly, the 40 m west and 40 m east will cancel out. Even though the object did travel, it ended up in the same position at the end of its journey as it was in its beginning position. That is why the objects displacement is 0 m

scoundrel [369]3 years ago

6 0

80,000 Is the answer I believe

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Jamie decides to drop an egg that has a mass of 345 g off the top of a 8.2 m building. How fast will it be falling right before

zlopas [31]

Answer:

13 m/s

Explanation:

Given:

Δy = 8.2 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (8.2 m)

v = 12.7 m/s

Rounded to two significant figures, the speed is 13 m/s.

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3 years ago

WILL GIVE BRAINLIEST DON'T KNOW IF I SPELLED IT WRONG BUT U KNOW WHAT I MEAN:)

eimsori [14]

Answer: well jabberwocky means like happy or like good vs evil

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Definition: invented or meaningless language; nonsense.

Explanation:

3 0

3 years ago

Read 2 more answers

In what way is Speed related to Kinetic Energy?

Papessa [141]

Answer:

Explanation:

In order to find kinetic energy, you need information about the velocity. Speed is the magnitude of the velocity.

Kinetic energy is 1/2mv^2

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8 0

3 years ago

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

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis.At

timurjin [86]

Answer:

a) The initial velocity of the green car is -13 m/s

b) The acceleration of the green car is - 2.25 m/s²

Explanation:

The equation for the position of objects moving in a straight line with constant acceleration is as follows:

x = x0 + v0·t + 1/2·a·t²

where:

x = position

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

If the velocity is constant, then a = 0 and x = x0 + v·t

a)The initial position of the red and green car is 0 m and 220 m respectively. We know that at 44.5 m the cars pass each other if the red car has a constant velocity of 20 km/h. So let´s find how much time it takes the cars to pass each other in this case:

The position of the red car is:

x = x0 + v·t

then:

0.0445 km = 0 km + 20 km/h · t

t = 0.0445 km/ 20 km/h = 8.0 s

We also know that if the red car has a velocity of 40 km/h, both cars pass each other at 76.6 m. So let´s find the time it takes the cars to reach that position using the equation for the red car:

0.0766 km = 0 km + 40 km/h · t

t = 0.0766 km / 40 km/h = 6.9 s

The position of the green car at t= 6.9 s and t = 8.0 s must be the same as the red car because both cars pass each other at those times.

Then, for the green car:

x = x0 + v0·t + 1/2·a·t²

0.0445 km = 0.220 km + v0 · 8.0 s + 1/2·a· (8.0 s)²

and

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·a· (6.9 s)²

Now we have a system of two equations with two unknowns.

Solving for "a" in the first equation

0.0445 km - 0.220 km - v0 · 8.0 s = 32 s²·a

(-0.176 km - v0 · 8.0 s) / 32 s² = a

Replacing a = (-0.176 km - v0 · 8.0 s) / 32 s² in the second equation and solving for v0:

0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·((-0.176 km - v0 · 8.0 s)/32 s²)·(6.9 s)²

-0.143 km = v0 · 6.9 s - 0.74(0.176 km + v0 · 8.0 s)

-0.143 km = v0 · 6.9 s - 0.130 km - v0 · 5.9 s

-0.143 km + 0.130 km = v0 · 6.9 s - v0 · 5.9 s

-0.013 km = 1 s · v0

v0 = -13 m/s

b) The acceleration of the green car is:

a = (-0.176 km - v0 · 8.0 s) / 32 s²

a = (-0.176 km - (-0.013 km/s) · 8.0 s) / 32 s² = -2.25 m/s²

3 0

3 years ago