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dexar [7]
3 years ago
12

Answer meeeeeeeeeeeeeee

Physics
2 answers:
Katarina [22]3 years ago
7 0
︎option A is the right one
konstantin123 [22]3 years ago
5 0

Answer:

option A is correct because air friction is greater than gravity

Explanation:

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Which organism are most likely to die from competition
Nataly_w [17]

Answer:

Intraspecific competition can serve as a regulator for population size. If a particular source of prey, or abiotic habitat feature is not readily available, then competition for the ones that are will be heavy. If the requirements are scarce enough, this will cause the population to remain stable, or decrease.

8 0
3 years ago
Read 2 more answers
1. A student gathered two boxes of the same size made of different materials: glass and clear plastic. She placed them on a wind
Kobotan [32]

Answer:

1.  to relate the type of box material to the temperature of the air within the box

2. Question is incomplete

3. scatterplot

Explanation:

1. The only thing done differently in this experiment is the type of material used in making the boxes, hence the experiment must be about that. Before proceeding to answering this question, we must have this at the back of our minds.

We can gather from the experiment that the boxes are of the same size and were subjected to sunlight for an hour (the same time duration for both). Hence, the temperature of the air inside the box will only be affected by the type of material the box is made of since the boxes have the same size and were subjected to sunlight for the same duration.

From the options provided, the best description for this experiment is; to relate the type of box material to the temperature of the air within the box.

2. The question is incomplete. The value for speed/velocity needed to calculate the average time is missing.

However, the formula needed here is velocity = distance ÷ time

3. There are two variables in this experiment; distance and time

The type of graph that shows two variables on it (of the options provided) is the scatterplot.

8 0
4 years ago
Pls help A ball rolls horizontally off a 100-meter-tall cliff at 40 meters per second. How far does the ball travel horizontally
svet-max [94.6K]

First find the time it takes for the ball to reach the ground using the vertical component of its position vector:

y=y_0+v_{0y}t+\dfrac12a_yt^2

\implies0=100\,\mathrm m+\dfrac12\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2

\implies t=4.52\,\mathrm s

Meanwhile, the horizontal component of the ball's position vector is

x=x_0+v_{0x}t+\dfrac12a_xt^2

\implies x=\left(40\,\dfrac{\mathrm m}{\mathrm s}\right)t

After about 4.52 s, the ball has traveled a horizontal distance of

x=\left(40\,\dfrac{\mathrm m}{\mathrm s}\right)(4.52\,\mathrm s)=180.8\,\mathrm m

which you would round to 200 m, so the answer is B.

8 0
3 years ago
A pulling force is called what?<br> A. normal<br> B. tension<br> C. balanced<br> D. compression
joja [24]
<span>There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)</span>
3 0
4 years ago
Read 2 more answers
A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b
Artist 52 [7]

<u>Answer:</u>

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

    Distance traveled during time t = 4 cm

    Initial velocity = u m/s

   Acceleration = 0 m/s^2

So 4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}

Now at time 2t,

  S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm

  So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

  Distance traveled during time t = 4 cm

    Initial velocity = 0 m/s

   Acceleration = -g m/s^2

4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}

At time 2t,

     S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm

 So vertical distance traveled in time 2t = 16 cm to the bottom

 

6 0
3 years ago
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