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Y_Kistochka [10]

3 years ago

11

How can you make a solution saturated

1 answer:

Step2247 [10]3 years ago

5 0

Answer: The answer is B. Add more solute (took test)

Explanation:

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If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

1 year ago

Four forces are exerted on a disk of radius R that is free to spin about its center, as shown above. The magnitudes are proporti

Dmitry_Shevchenko [17]

The given magnitude of forces of F₁ = F₄, F₂ = F₃, F₁ = 2·F₂, give the

forces that exert zero net torque on the disk as the options;

(B) F₂

(D) F₄

<h3>How can the net torque on the disk be calculated?</h3>

The given parameters are;

F₁ = F₄

F₂ = F₃

F₁ = 2·F₂

Therefore;

F₄ = 2·F₂

In vector form, we have;

$\vec{F_4} = \mathbf{\frac{\sqrt{3} }{2} \cdot F_4 \cdot \hat i - 0.5 \cdot F_4 \hat j}$

$\vec{F_2} = \mathbf{ -F_2 \, \hat j}$

Clockwise moment due to F₄, M₁ = $-0.5 \times F_4 \, \hat j \times \dfrac{R}{2}$

Therefore;

$M_1 =- 0.5 \times 2 \times F_2 \, \hat j \times \dfrac{R}{2} = \mathbf{ -F_2 \, \hat j \times \dfrac{R}{2}}$

Counterclockwise moment due to F₂ = $-F_2 \, \hat j \times \dfrac{R}{2}$

Given that the clockwise moment due to F₄ = The counterclockwise moment due to F₂, we have;

Two forces that combine to exert zero net torque on the disk are;

F₂, and F₄

Which are the options; (B) F₂, and (D) F₄

Learn more about the resolution of vectors here:

brainly.com/question/1858958

4 0

2 years ago

Find the noise level of a sound having an intensity of 1.5x10^-14W/m^2 given I0=10^12 W/m2

Llana [10]

Answer:

Noise level will be -18.2 watt

So option (b) will be correct answer

Explanation:

We have given sound intensity $I=1.5\times 10^{-14}w/m^2$

And threshold intensity $I_0=\times 10^{-12}w/m^2$ ( in question it is given as $10^{12}w/m^2$ but its standard value is $10^{-12}w/m^2$ )

Now noise level $=10log\frac{I}{I_0}=10log\frac{1.5\times 10^{-14}}{10^{-12}}=10log0.015-18.23$

So the noise level will be -18.2

So option (b) will be correct answer

5 0

3 years ago

A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

Rashid [163]

Answer:

$v_{max} = |d|\cdot \sqrt{\frac{k}{m} }$

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$k \cdot d^{2} = m\cdot v^{2}$

$v_{max} = |d|\cdot \sqrt{\frac{k}{m} }$

5 0

3 years ago

Please help <br> I need it queekly

olga_2 [115]

Answer:

it is cheap to use as many fuels are expensive and do the same as destroying our environment

that's according to my consciousness

6 0

3 years ago