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3 years ago

TRUE OR FALSE: The following drops were most likely dropped from a 90 degree angle.

Physics

2 answers:

horsena [70]3 years ago

7 0

Answer:

True

Explanation:

If it weren't from a 90 degree angle then the circle would be a bit more oval shaped

qaws [65]3 years ago

3 0

Answer:

I think its true. Hope this helps.

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If a persons skin is dry she will experience more of a shock

NeX [460]

Answer:

add lotion

Explanation:

7 0

2 years ago

Read 2 more answers

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$

$\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}$

Removing $y_o$ and dividing by t (t different of zero)

$\displaystyle 0=v_osin\theta-\frac{gt}{2}$

Then we find the total flight as

$\displaystyle t=\frac{2v_osin\theta}{g}$

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

$\boxed{t=24.5/2=12.25\ seconds}$

4 0

3 years ago

Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los A

WARRIOR [948]

Answer:

a.Beth

b.2232 s

Explanation:

We are given that

Distance,d=400 mi

Speed of Alan,v=45 mph

Speed of Beth,v'=55 mph

a.Time = $\frac{distance}{speed}$

Using the formula

Time taken by Alan= $\frac{400}{45}=8.89 hr$

Time taken by Beth= $\frac{400}{55}=7.27hr$

Alan will reach San Francisco at 4:53 PM

Beth will reach San Francisco at 4:16 PM

Beth will reach before Alan.

b.Difference between time=8.89-7.27=1.62 hr

t=1.62 hr

1.62-1=0.62 hr

0.62 hr= $0.62\times 60\times 60=2232 s$

Hence, Beth has to wait 2232 s for Alan to arrive .

6 0

3 years ago

A body is electrically neutral. Does it mean that it has no charge?<br>

vova2212 [387]

Answer:

If a body is electrically neutral it means it has no net charge.

Explanation:

Because it has the same number if protons as it does electron , which are opposite charges that offset eachother.

7 0

3 years ago

I need help on this.

lana [24]

Answer:

The speed change during the 45-minute trip is 20[mph]

Explanation:

When we see the speed at the 45 minutes this is 20 [mph] and at the 0 minutes the speed is 0 [mph].

Therefore the change is (20 - 0) = 20 [mph]

In the attached image we can see the different figures. In fig 1 we can see the bicycle's speed after 10 minutes when the speed becames constant.

In the fig. 2 we can find the graph when the biker stopped at 30 minutes and took a 15-minute break.

Figures 3 and 4, show the differences when a horizontal line is traced on a position vs time graph, and when the horizontal line is traced in a speed vs time graph.

For fig 3 we can conclude that the body is not moving therefore there is no velocity or acceleration. And for the fig 4, we can realize that the area under the horizontal line represents a displacement during the respective interval of time.

3 0

3 years ago