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3 years ago

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Physics

2 answers:

mixas84 [53]3 years ago

8 0

Answer:

nice!!! but did you know that geico can save you 15% or more on car insurance!

Explanation:

ANTONII [103]3 years ago

8 0

Answer:

yes bestieee

Explanation:

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What is the best description of a mechanical wave?

gizmo_the_mogwai [7]

Answer:

A mechanical wave is a wave that is an oscillation of matter, and therefore transfers energy through a medium. While waves can move over long distances, the movement of the medium of transmission—the material—is limited. Therefore, the oscillating material does not move far from its initial equilibrium position.

Explanation:

8 0

3 years ago

The initial speed of a body is 7.1 m/s. What is its speed after 2.23 s if it accelerates uniformly at 2.64 m/s 2 ? Answer in uni

Nana76 [90]

13.0m/s

1.2m/s

Explanation:

Given parameters:

Initial speed of the body = 7.1m/s

time taken = 2.23s

Acceleration = 2.64m/s²

Unknown:

Final speed = ?

Solution:

Acceleration is the rate of change of velocity with time.

a = $\frac{V - U}{T}$

a = acceleration

V = final speed

U = initial speed

T = time taken

Input the variables and solve for V;

2.64 = $\frac{V - 7.1}{2.23}$

V - 7.1 = 5.9 expression 1

V = 5.9 + 7.1 = 13.0m/s

Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;

from expression 1;

V - 7.1 = -5.9

V = -5.9 + 7.1 = 1.2m/s

learn more:

Acceleration brainly.com/question/3820012

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3 years ago

The study of the evolution and the mind using principles of natural selection is evolution

Rus_ich [418]

this is evolutionary psychology

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3 years ago

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h

Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2$

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft$