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3 years ago

Ponds and lakes are part of marine ecosystem. True or false ?

Physics

2 answers:

finlep [7]3 years ago

8 0

Answer:

True

Explanation:

because they can hold marine organism inside

nevsk [136]3 years ago

7 0

False because they are fresh water

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What is the slope of the line plotted below? (-2,-1) (4,2) A. 1 B. 2 C. -0.5 D. 0.5

Dmitry [639]

Answer: D. 0.5

Explanation:

The slope formula is y2-y1/x2-x1.

2-(-1)/4-(-2) = 2+1/4+2

2+1/4+2 = 3/6 = 1/2

1/2 = 0.5

The slope is 1/2, or 0.5.

8 0

3 years ago

The body weighing 8kg moves in a straight line uniformly accelerated with an acceleration of 3m / s? on a horizontal surface, un

Usimov [2.4K]

Answer:

pls explain this to me

Explanation:

I will not answer it

3 0

3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

What is a gravitational field and how its strength be measured

Yakvenalex [24]

A gravitational field is the field generated by a massive body, that extends into the entire space. Every object with mass m experiences a force F when immersed in a gravitational field. The intensity of the force is equal to
$F= \frac{GM}{r^2} m$
where $G=6.67 \cdot 10^{-11} m^3 Kg^{-1} s^{-2}$ is the gravitational constant, M is the mass of the source of the field (e.g. the mass of a planet), and r is the distance between the object and the source of the field. The force is always attractive.

A possible way to measure the intensity of a gravitational field is by measuring the acceleration a of the object immersed in this field. In fact, for Newton's second law we have:
$F=ma$
but since
$F= \frac{GM}{r^2} m$
we can write
$a = \frac{GM}{r^2}$
Therefore, by measuring the acceleration of the object, we also measure the intensity of the field.

5 0

3 years ago

A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied.

Morgarella [4.7K]

a) The car moved 73.333 feet when its speed has been reduced to 30 miles per hour.

b) The car moved 117.333 feet when its speed has been reduced to 15 miles per hour.

c) The curve is described by $v = \sqrt{4356 -33\cdot \Delta s}$ , where $v$ , in feet per second, and $\Delta s$ , in feet, and the graph is included in the image below.

<h3>Procedure - Determination of current positions for a car that decelerates uniformly</h3>

In this question we must use the following <em>kinematic</em> formula to determine the deceleration rate ( $a$ ), in feet per square second, experimented by the vehicle:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s}$ (1)

Where:

$\Delta s$ - Travelled distance, in feet.
$v_{o}$ - Initial speed, in feet per second.
$v$ - Final speed, in feet per second.

If we know that $\Delta s = 132\,ft$ , $v_{o} = 66\,\frac{ft}{s} \left(45\,\frac{mi}{h}\right)$ and $v = 0\,\frac{ft}{s}$ , the deceleration rate experimented by the car is:

$a = \frac{\left(0\,\frac{ft}{s}\right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot (132\,ft)}$

$a = -16.5\,\frac{ft}{s^{2}}$

<h3>a) The distance traveled by the car when speed has been reduced to 30 miles per hour</h3>

By using (1) and knowing that $v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)$ , $v = 44\,\frac{ft}{s}\left(30\,\frac{mi}{h} \right)$ and $a = -16.5\,\frac{ft}{s^{2}}$ , we find the distance travelled by the vehicle:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (1b)

$\Delta s = \frac{\left(44\,\frac{ft}{s} \right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot \left(-16.5\,\frac{ft}{s^{2}} \right)}$

$\Delta s = 73.333\,ft$

The car moved 73.333 feet when its speed has been reduced to 30 miles per hour. $\blacksquare$

<h3>b) The distance traveled by the car when speed has been reduced to 15 miles per hour</h3>

By using the same approach used in part (a) and knowing that $v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)$ , $v = 22\,\frac{ft}{s}\left(15\,\frac{mi}{h} \right)$ and $a = -16.5\,\frac{ft}{s^{2}}$ , we find the distance travelled by the vehicle:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(22\,\frac{ft}{s} \right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot \left(-16.5\,\frac{ft}{s^{2}} \right)}$

$\Delta s = 117.333\,ft$

The car moved 117.333 feet when its speed has been reduced to 15 miles per hour. $\blacksquare$

<h3>c) Graph of the distance vs speed</h3>

In this case we must use (1) in the following form to obtain every speed associated with each travelled distance:

$v = \sqrt{v_{o}^{2}+2\cdot a\cdot \Delta s}$ (1c)

If we know that $v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)$ and $a = -16.5\,\frac{ft}{s^{2}}$ , then we have the following formula and respective graph.

$v = \sqrt{4356 -33\cdot \Delta s}$ $\blacksquare$

And the graph is presented in the image attached below. $\blacksquare$

To learn more on uniform accelerated motion, we kindly invite to check this verified question: brainly.com/question/12920060

6 0

2 years ago