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2 years ago

6

It's five miles from Tim's house to Rita's house. Roughly how long is this distance in feet?

2 answers:

Ber [7]2 years ago

5 0

The answer is 26400

sergey [27]2 years ago

3 0

Answer:

26400 ft

Explanation:

that would be my answer

hope this helps!!! :)

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For which pair of launch angles will two identical projectiles have equal ranges?. A. 19.24°, 80.54°B. 16.42°, 74.58°C. 60.23°,

AlladinOne [14]

The kinematic equations of motion that apply here are<span>y(t)=votsin(θ)−12gt2</span>and<span>x(t)=votcos(θ)</span>Setting y(t)=0 yields <span>0=votsin(θ)−12gt2</span>. If we solve for t, we obtain, by factoring,<span>t=<span>2vsin(θ)g</span></span>Substitute this into our equation for x(t). This yields<span>x(t)=<span><span>2v2cos(θ)sin(θ)</span>g</span></span><span>This is equal to x=<span><span>v^2sin(2θ)</span>g</span></span>Hence the angles that have identical projectiles are have the same range via substitution in the last equation is C. <span> 60.23°, 29.77° </span>

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3 years ago

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A train travels due south at 25 m/s (relative to the ground) in a rain that is blown toward the south by the wind. The path of e

olasank [31]

Answer:

Explanation:

Given

Train travels towards south with a velocity if $v_t=25\ m/s$

Rain makes an angle of $\theta =66^{o}$ with vertical

If an observer sees the drop fall perfectly vertical i.e. horizontal component of rain velocity is equal to train velocity

suppose $v_r$ is the velocity of rain with respect to ground then

$v_r\sin\theta =v_t$

$v_r\times \sin (66)=25$

$v_r=27.36\ m/s$

Therefore velocity of rain drops is 27.36 m/s

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3 years ago

Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T

olya-2409 [2.1K]

The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

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2 years ago

What happens during the fourth stage of technological design that is not necessary during the sixth or seventh stages of scienti

riadik2000 [5.3K]

<span>A design is remodeled after analysis and tested again.</span>

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2 years ago

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Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from

In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 $m/s^2$

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

$v = u + gt$

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

$49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs$

The arrow was in flight for 3.8 secs

6 0

3 years ago