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nika2105 [10]
3 years ago
12

Answer any of these questions

Physics
1 answer:
Nataly [62]3 years ago
7 0

1 to 3 S, for the bottom question, and 1 s for the top questions (guessing, sorry)

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PLS HELP!!! WILL GIVE BRAINLIEST
grandymaker [24]

Answer:

did you ever get the answer

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The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
What work is done by a system on its environment when an ideal gas is maintained at a constant pressure of 200 newtons/meter2 du
Crank

<u>Answer:</u>

Work done done by a system on its environment is W=40 Joules

<u>Explanation:</u>

<u>Given:</u>

Ideal gas is maintained at a constant pressure P=200 \frac{N}{m^{2}}

Change in Volume d v= -0.2 m^{3}

<u>To find:</u>  

Work done by a system during an Isobaric Process

<u>Solution:</u>  

In a Isobaric Process, constant pressure is maintained, P=200 \frac{N}{m^{2}}

According to the formula,

work done  \bold{W=P \times d v}

Substitute the values of pressure and change in volume in the above formula,

=200 \times -0.2

=40 \text { Joules }

<u>Result:</u>  

Work done done by a system W= -40 Joules

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Answer:

what is this I don't get a thing

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A. Community<br> B. Organism<br> C. Population <br> D. Individual
dimulka [17.4K]

Answer:

i think its B or D

Explanation:

its Individual or Organism

3 0
3 years ago
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