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Pavlova-9 [17]
3 years ago
8

A sample of hydrated calcium sulfate has a mass of 20.88 g. After it is heated, it has a mass of 16.51 g. What is the percentage

by mass of water in the hydrate? Use the formula mc010-1.jpg 4.37% 20.9% 26.5% 43.7%
Physics
2 answers:
STALIN [3.7K]3 years ago
6 0
We know that when you heat up calcium sulfate all the water has evaporated from it. That means that the mass of water that was initially in the sulfate is the difference in mass of the sulfate before and after the heating.
m{_H__20}=20.88-16.51=4.37g
In order to <span>get a percentage by mass we simply divide this number by the initial amount of sulfate:
</span>\%_{H__2}O=\frac{4.37}{20.88}=20.93\%<span>
</span>
nexus9112 [7]3 years ago
6 0
The answer is B) 20.9%
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A small cube of metal measures 19.0 mm on a side and weighs 79.6 g. What is the density of the metal in g/cm3?
Nady [450]

Answer:

density of cube =11.605 g/cm³

Explanation:

density of a substance is the mass per unit volume of that substance.

the density of a substance = \frac{mass}{volume}

volume of a cube = l³,

l = 19.0mm , lets convert mm to cm

1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm

length of cube =1.9cm

volume of cube = 1.9³

density of cube = \frac{79.6}{1.9^{3} }

density of cube =11.605 g/cm³

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4 years ago
A seesaw pivots as shown in below. What is the net torque about the pivot point?
rodikova [14]

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2.3 Nm clockwise

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The net torque is 2.3 Nm clockwise.

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Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

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15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

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r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

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1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

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r_x=v_0t

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3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

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