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motikmotik
3 years ago
15

Solve: A car travels 2 km North , 10 km East, then 3 km West. pythagorean theorem

Physics
1 answer:
Evgen [1.6K]3 years ago
5 0

________b_____ 7 km east

|

| 2km north.

|a

|

°

pythagorean theorem : ✓a² + b² = c²

c² = a² + b² = 4 + 49 = 53

c = ✓53 km

displacement = c = ✓53 km

distance = 10 + 3 + 2 = 15 km

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A mango fruit drops from the top of it's tree which is 40 m. How long does it takes to reach the ground
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4 0
3 years ago
A vector has a component of 10 m in the x-direction, a component of 10 m in the y-direction, and a component of 5 m in the z dir
Kobotan [32]

The magnitude of this vector is 15

A vector is a quantity or phenomenon that has two independent properties: magnitude and direction. The term also denotes the mathematical or geometrical representation of such a quantity. Examples of vectors in nature are velocity, momentum, force, electromagnetic fields, and weight.

The magnitude of a vector formula is used to calculate the length for a given vector (say v) and is denoted as |v|. So basically, this quantity is the length between the initial point and endpoint of the vector.

Let vector be = a

component of vector in x direction = 10 i

component of vector in y direction = 10 j

component of vector in z direction = 5 z

vector a = 10 i + 10 j + 5 z

magnitude of vector a = |a| = \sqrt{10^{2} +10^{2} + 5^{2}    }

                                             = 15

To learn more about vector here

brainly.com/question/24256726

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8 0
2 years ago
A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises
atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s²  * t²

-2.0 m = -4.9 m/s²  * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t      where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t²           y = 1.5 m       y0 = 0 m   t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

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