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1 year ago

What is the concentration of a solution if 65 g of sugar is dissolved into 200 ml of water?

Physics

1 answer:

DIA [1.3K]1 year ago

5 0

Answer:

the concentration in g/L 325 while in molarity is 0.6398(M).

Explanation:

hope this helps

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A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential

nlexa [21]

Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

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2 years ago

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Brainlist nd 20 points

nadezda [96]

Answer:

I'm not sure but I think it's 35-39

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2 years ago

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An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

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2 years ago

A star has an absolute magnitude of 4 and a surface temperature of 5,000 degrees C. According to the HR diagram, list the type o

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2 years ago

A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.

Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

$a=\frac{v-u}{t}$

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

$a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2$

And now we can calculate the force exerted on the running back, by using Newton's second law:

$F=ma=(100 kg)(-10 m/s^2)=-1000 N$

so, the magnitude of the force is 1000 N.

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2 years ago

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