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3 years ago

Joe rides his bicycle in a straight line for 2 hour with an average velocity of 13 km/h south, how far has he ridden?

Physics

1 answer:

dmitriy555 [2]3 years ago

8 0

Answer:

ans is 3.125 km

i hope this is correct

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Can u Anwser this Plzzzz

nata0808 [166]

Answer:

-0. 75m/s^2

Explanation:

use formula of acceleration

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3 years ago

In case of collision of objects in two dimensions which statement is true after the collision?

Grace [21]

The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved.

EXPLANATION :

Before coming into any conclusion, first we have to understand the law of conservation of momentum.

As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.

Hence, during any type of collision, the total momentum is always conserved.

Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.

5 0

3 years ago

Read 2 more answers

Why was Earth very hot at the beginning of its formation? Select all that apply.

Sladkaya [172]

Answer:

heat came from the supernova that led to the formation of solar system

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3 years ago

Mester Exam 1 11 of 35

frez [133]

Please show picture of diagrams

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3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$