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2 years ago

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An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o

f the apple.

1 answer:

Crazy boy [7]2 years ago

3 0

<span>umm it gives 1/v + 1/20 = -1/10 and v = -20/3 cm. Magnification = |v/u|=1/3.</span>

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The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted val

yaroslaw [1]

Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

<em>= |174.5 - 173.2|</em>

<em>= |1.3|</em>

<em>Percentage error = (Error / Theoretical value) × 100</em>

<em>= (1.3 / 173.2)×100</em>

<em>= 0.75 %</em>

∴ Percentage error in the reading is 0.75%

4 0

2 years ago

Match the correct term with each part of the wave

Misha Larkins [42]

6 . . . . . a crest
7 . . . . . the amplitude
8 . . . . . the wavelength
9 . . . . . a trough

6 0

3 years ago

What two bodily functions are increased by a warm up

drek231 [11]

1) Blood flow: increases during warming improving muscle and joint elasticity. This decreases the possibility of having an injury.

2) Body temperature: This causes the cellular metabolism to increase. It also causes vasodilatation that allows a greater supply of oxygen and nutrients.

6 0

2 years ago

Which of the following is a false statement about dispersion forces? View Available Hint(s) Which of the following is a false st

icang [17]

Explanation :

Dispersion forces are also known as London dispersion forces. It is the weakest force. Also, it is the part of the Van der Waals forces.

(1) This force is exhibited by all atoms and molecules.

(2) These forces are the result of the fluctuations in the electron distribution within molecules or atoms. Due to these fluctuations, the electric field is created. The magnitude of this force is explained in terms of Hamaker constant 'A'.

(3) Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. When electrons are more concentrated in a place, instantaneous dipoles formed.

(4) Dispersion force magnitude depends on the amount of surface area available for interactions. If the area increases, the size of the atom also increase. As a result, stronger dispersion forces.

So, the false statement is "Dispersion forces always have a greater magnitude in molecules with a greater molar mass".

4 0

2 years ago

Solve for work when

BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.

<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

$\boxed{\sf{\bold{W = F \times s}}}$

If the Angle Effect on Work

$\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}$

With the following condition:

W = work that done by object (J)
F = force that applied (N)
s = shift or distance (m)
$\sf{\theta}$ = angle of elevation (°)

<h3>Solution</h3>

We know that :

F = force that applied = $\sf{1.41 \times 10^4}$ N
s = shift or distance = 84.9 m
$\sf{\theta}$ = angle of elevation = 45°

What was asked ?

W = work that done by object = ... J

Step by step :

$\sf{W = F \times s \times \cos(\theta)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 59.8545 \sqrt{2}}$

$\boxed{\sf{W \approx 84.65 \: J}}$

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0

8 months ago