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3 years ago

What is the connection between the x- and y-motions of a projectile?

Physics

2 answers:

Vikentia [17]3 years ago

7 0

Answer:

D. height

Explanation:

Sunny_sXe [5.5K]3 years ago

5 0

Answer c, velocity would be the answer.

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10. What is Newton's 3rd Law?

charle [14.2K]

Answer:

A. For every action there is an equal and opposite reaction.

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2 years ago

Read 2 more answers

Just the the moon seems to shine and the two below it.

katrin [286]

The first one is revolves

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3 years ago

A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>2) m>s2, where s is in meters. De

Len [333]

Answer:

v=1.295

Explanation:

What we are given:

a=5÷(3s^(1/3)+s^(5/2)) m/s^2

Start by using equation a ds = v dv

This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

$\int\limits^a_b {x} \, dx$

a=2

b=1

x=5÷(3s^(1/3)+s^(5/2)) m/s^2

dx=dv

Integrate the left side the standard method.

$\int\limits^a_b {x} \, dx$

a=v

b=0

dx=dv

Integrating

=v^2/2

Use Simpson's rule for the right site.

$\int\limits^a_b {x} \, dx$

a=b

b=a

x=f(x)

f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)

If properly applied. you should now have the following equation:

v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]

=0.8376

Solve for v.

v=1.295

5 0

3 years ago

A student conducts an experiment in which an object travels across a horizontal surface while for 2 s a net force is applied to

erma4kov [3.2K]

Answer:

D. No, because the student needs to know the direction that the force is applied

Explanation:

The change in momentum depends on the direction of the force as well as its magnitude. Since the graph only supplies force magnitude information, it is insufficient to allow the student to calculate the change in momentum.

3 0

3 years ago

If, for a given velocity, the maximum range is at a projection angle of 45, then there must be equal ranges for angles above and

tiny-mole [99]

Explanation:

The range R of a projectile is given by

$R = \frac{v_0^2}{g} \sin 2\theta$

The maximum range $R_{max}$ occurs when $\sin 2\theta = 1\:\text{or}\:\theta = 45°$ . Let $\alpha$ be the angle above or below 45°. Now let's look at the ranges brought about by these angle differences.

Case 1: Angle above 45°

We can write the range as

$R_+ = \dfrac{v_0^2}{g} \sin 2(45° + \alpha)= \dfrac{v_0^2}{g} \sin (90° + 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha + \cos 90° \sin 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(1)$

Case 2: Angle below 45°

We can write the range as

$R_- = \dfrac{v_0^2}{g} \sin 2(45° - \alpha)= \dfrac{v_0^2}{g} \sin (90° - 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha - \cos 90° \sin 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(2)$

Note that the equations (1) and (2) are identical. Therefore, the ranges are equal if they differ from 45° by the same amount.

3 0

3 years ago