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2 years ago

6

Choose all the answers that apply. Models are used to observe phenomena which _____. happen too slowly} happen too quickly} are

too large} are too small } are too dangerous

2 answers:

bazaltina [42]2 years ago

8 0

your answer is phenomena which are too large

Paraphin [41]2 years ago

4 0

Answer:

All of these answer are used in models.

Explanation:

The use of models is fundamentals tool in science. Scientists used the models to observe phenomena which happen too slowly, or happen too quickly or may to large or small process and how to extent it may dangerous for firsthand. They used different model depending upon whether which type of question is investigating.

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How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half

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Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

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- the voltage is decreased to half of its original value: $V'=\frac{V}{2}$

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2 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:

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

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

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