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Scilla [17]
2 years ago
7

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]2 years ago
5 0

Answer:

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

Explanation:

hope this helps

You might be interested in
I need answers and solvings to these questions​
den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) 15.7 rad/s^2

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f is the final angular velocity

\omega_i is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

\omega_i = 0 (since it starts from rest)

\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s is the final angular velocity

t = 2 s

Substituting, we find

\alpha = \frac{31.4-0}{2}=15.7 rad/s^2

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

a=-\omega^2 x

where

a is the acceleration

x is the displacement

\omega is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

a=-100 x

which implies that

\omega^2 = 100

And so

\omega = \sqrt{100}=10 rad/s

Also, the angular frequency is related to the frequency f by

f=\frac{\omega}{2\pi}

Therefore, the frequency of this simple harmonic oscillator is

f=\frac{10}{2\pi}=1.6 Hz

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

mg=kx

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5

The angular frequency of the spring is given by

\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz

And therefore, its period is

T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the hose, with r_1 = 5 mm being the radius of the hose

v_1 = 4 m/s is the speed of the petrol in the hose

A_2 = \pi r_2^2 is the cross-sectional area of the nozzle, with r_2 being the radius of the nozzle

v_2 = 16 m/s is the speed in the nozzle

Solving for r_2, we find the radius of the nozzle:

\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm

So, the diameter of the nozzle will be

d_2 = 2r_2 = 2(2.5)=5.0 mm

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

\frac{F_1}{A_1}=\frac{F_2}{A_2}

where

F_1 is the force that must be applied to the small piston

A_1 = \pi r_1^2 is the area of the first piston, with r_1= 2 cm being its radius

F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N is the force applied on the bigger piston (the weight of the car)

A_2 = \pi r_2^2 is the area of the bigger piston, with r_2= 15 cm being its radius

Solving for F_1, we find

F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0
3 years ago
A student notices that an inflated balloon gets larger when it is warmed by a lamp. Which best describes the mass of the balloon
nirvana33 [79]
It gets larger because
well let me give you an example
so today in class we looked at a lava lamp with wax inside and there was a lightbulb at the bottom.
we watched as the wax floated up because the molecules inside the wax spreads out and makes the wax less dense.
the wax floats up because (which is related to the balloon getting bigger) the wax is getting less dense and the particles get bigger which ALSO makes the wax less dense.
hope this helps and hope you can relate it to your problem! say thanks if I did help AT ALL! :)
7 0
3 years ago
Read 2 more answers
A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .
sp2606 [1]
How much work in J does the string do on the boy if the boy stands still? 

<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>

<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>

<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>

<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>

<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
6 0
3 years ago
Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l
Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1       (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2       (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2)     (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5    (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2        (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067}  \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

   = 0.5 - 0.9931 × 0.340

   = 0.5 - 0.338

   = 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0
2 years ago
Three balls with different masses are shown below.
algol [13]

Answer:

$10 a g

Explanation:

6 0
3 years ago
Read 2 more answers
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