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3 years ago

Which is a chemical property that can be used to identify calcium carbonate?

Physics

2 answers:

daser333 [38]3 years ago

8 0

Answer:

could someone at least give one answer instead of a choice between two cause that is a 50/50 chance to get it correct on a test!!

Explanation:

Anna71 [15]3 years ago

5 0

(D) reacts with acids by producing gas bubbles
Or
(A) White Color

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Gauss’ law: a. Relates the surface charge density to the electric field.b. Relates the electric field at points on a closed surf

Mamont248 [21]

Answer:

b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

Explanation:

Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

8 0

3 years ago

A crate slides down a ramp that makes a 20o angle with the ground. To keep the crate moving at a steady speed, Paige pushes back

cupoosta [38]

Answer:

-223.64684 J

Explanation:

F = Force that is applied to the crate = 68 N

s = Displacement of the crate = 3.5 m

$\theta$ = Angle between the force and displacement vector = (180-20)

Work done is given by

$W=Fscos\theta\\\Rightarrow W=68\times 3.5\times cos(180-20)\\\Rightarrow W=-223.64684\ J$

The work that Paige does on the crate is -223.64684 J

6 0

3 years ago

I attempted to answer and got 0m, please explain how to get to the answer.

mafiozo [28]

The maximum height to which the ball attain before falling back down is 1147.96 m

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

Initial velocity (u) = 150 m/s
Final velocity (v) = 0 m/s (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =?

<h3>How to determine the maximum height </h3>

The maximum height reached by the ball can be obtained as illustrated below:

v² = u² – 2gh (since the ball is going against gravity)

0² = 150² – (2 × 9.8 × h)

0 = 22500 – 19.6h

Collect like terms

0 – 22500 = –19.6h

–22500 = –19.6h

Divide both side by –19.6

h = –22500 / –19.6

h = 1147.96 m

Thus, the maximum height reached by the ball is 1147.96 m

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

5 0

1 year ago

The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

leonid [27]

Answer:

The magnitude of the force is $F_{net}= 1.837 *10^4N$

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

At t = 0 , $\theta = 20^o$

The rate at which the angle increases is $w = 2 \ ^o/s$

Converting this to revolution per second $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The length of the rope is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , The tension on the rope T = 18 kN

Mass of the para-sailor is $M_p = 75kg$

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

Now the derivative of displacement is velocity

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

represents the velocity, again the derivative of velocity gives us acceleration

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now to the time when the rope made angle of 30° with the water

generally angular velocity is mathematically represented as

$w = \frac{\Delta \theta}{\Delta t}$

Where $\theta$ is the angular displacement

Now considering the interval between $20^o \ to \ 30^o$ we have

$2 = \frac{30 -20 }{t -0}$

making t the subject

$t = \frac{10}{2}$

$= 5s$

Now at this time the displacement is

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity is

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

The linear acceleration is

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally radial acceleration is mathematically represented by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Generally angular acceleration is mathematically represented by

$\alpha_t = r \theta'' + 2 r' \theta '$

Now $\theta '' = \frac{d (0.0349)}{dt} = 0$

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The net resultant acceleration is mathematically represented as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the direction of the is acceleration is mathematically represented as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The force on the para-sailor along y-axis is mathematically represented as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The force on the para-sailor along x-axis is mathematically represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The net resultant force is mathematically evaluated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The direction of the force is

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

3 years ago

Which of the following requires constant agonist-antagonist muscle contraction

vodka [1.7K]

Answer: Dynamic balance

Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121

5 0

3 years ago