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2 years ago

11

Environment A is warm and gets lots of rain. Environment B is warmer and gets more rain year round. Identify Environment A.(2 po

ints)
Tundra

Rainforest

Desert

Swamp

1 answer:

zlopas [31]2 years ago

5 0

The Swamp
There are many websites that say the rainforest but the rainforest is warmer and gets rain year round and swamps are warm and gets lots of rain but not year round.

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What provides the force on the person in the passenger seat?

podryga [215]

The forces that make a passenger speed up, slow down, or
turn a curve are the same forces that have the same effect
on the driver and anybody else in the car.

-- Speeding up . . .

the back of the seat
    friction between the car seat and the seat of your pants

-- Slowing down . . .

the seat belt
      friction between the car seat and the seat of your pants

-- Turning away from a straight line . . .

      the seat belt
friction between the car seat and the seat of your pants
the door, or whatever or whomever you're leaning against

6 0

4 years ago

A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g

Sliva [168]

To solve for force, you need to get the product of mass and acceleration.
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

$a = \frac{vf-vi}{t}$

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.

So we can put it into our formula now:

$a = \frac{0m/s-50m/s}{20ms}$

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.

$20ms$ x $\frac{1s}{1000ms}$ = $\frac{20s}{1000} = 0.02s$

So your new time is 0.02s. Now we put this time into the formula:

$a = \frac{0m/s-50m/s}{0.02s}$
$a = \frac{-50m/s}{0.02s} = -2,500 m/ s^{2}$

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.

So now we have our acceleration. Now using this, we can get our force.

$F= ma$

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is $kg.m/ s^{2}$ but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.

$150g = \frac{1kg}{1,000g} = \frac{150kg}{1,000} = 0.150kg$

Our new mass is 0.150kg.

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 $m/ s^{2}$

Now that all our units match, we can put that into our formula:

$F= ma$
$F= (0.150kg)(-2,500m/s^{2})$
$F = -375kg.m/ s^{2} or -375N$

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.

4 0

4 years ago

On a summer afternoon, the sand on the beach can get very hot. When you step on the sand in bare feet, you can burn yourself.

8_murik_8 [283]

<span>Heat from the Sun is transferred to the sand without direct contact. This heat is then transferred to your feet by direct contact.</span>

8 0

3 years ago

Read 2 more answers

If you make multiple measurements of your height, you are likely to find that the results vary by nearly half an inch in either

Lisa [10]

Answer:

Height h= 1.7 m

Explanation:

Supposing we have to find height in meter.

1 feet = 0.3048 m

1 inch = 0.0254 m

Given that:

5 feet

= 5×0.3048

= 1.524 m

and 7 inch = 7×0.0254= 0.1778 m

Therefore total height of a man in meter

5 feet 7 inch = 1.5424+0.1778 =1.7 m

Height h= 1.7 m

8 0

3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time <em>t</em> is

$v_{y,f}=v_{y,i}-gt$

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction <em>θ</em> such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$

6 0

3 years ago