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2 years ago

What force types change the direction of motion

2 answers:

7 0

Forces affect how objects move. They may cause motion; they may also slow, stop, or change the direction of motion of an object that is already moving. Since force cause changes in the speed or direction of an object, we can say that forces cause changes in velocity. Remember that acceleration is a change in velocity. Let’s say an object is moving along a table on earth, suddenly the finite table ends, resulting in the object being present in the air, which means there is no normal contact force N to combat the force by gravity mg, which is why there is an acceleration downwards. This proves as a projectile motion since the direction of motion start changing from horizontal to vertical. Another example is one throwing an object up. It moves up and slows down, reaching its maximum point, leading to it starting to move downwards. This too is a change in motion.

Leno4ka [110]2 years ago

7 0

I believe an unbalanced force.

Unbalanced force is the change in direction of motion.

Here, look at the following definition: If two individual forces are of equal magnitude and opposite direction, then the forces are said to be balanced. An object is said to be acted upon by an unbalanced force only when there is an individual force that is not being balanced by a force of equal magnitude and in the opposite direction.

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The sun supplies about 1.0 kilowatt of energy for each square meter of surface area (1.0 kW/m^2 where a watt = 1 kJ/s) Plants pr

kaheart [24]

Answer:

0.092 %

Explanation:

The equation of the reaction can be computed as :

$12CO_2_{(g)} + 11H_2O_{(l)} \to C_{12}H_{22}O_{11} + 12O_{2_(g)}$

$\Delta H = 5645 \ kJ$

recall that; the number of moles = $\dfrac{mass}{molar \ mass}$

By applying the method of enthalpy of combustion for sucrose at the same time changing the time from hours to seconds, we can determine the total energy output.

i.e

$=\dfrac{0.20g \ of \ sucrose }{m^2 \ 3600 \ s}\times \dfrac{1 \ mol}{342.34 \ g}\times 5.645 kJ/mol$

$= 9.16 \times 10^{-4} \ kJ/m^2 s$

Given that the sun supplies about 1.0 kilowatt, to KJ/m² s, we have:

$1.0 \dfrac{kW}{m^2 }= 1.0 \dfrac{kJ}{m^2 s}$

Finally, the percentage of sunlight used to produce sucrose :

= $\dfrac{9.16 \times 10^{-4} \ kJ/m^2 \ s}{1.0 \ kJ/m^2 . s} \times 100\%$

= 0.092 %

6 0

3 years ago

What is the oxidation number of P in PF6^-

Hitman42 [59]

<span>F can only have oxidation number of -1
The overall compound has an oxidation number of -1
You have 6 Fs, so 6(-1) = -6
charge from F
X = oxidation number of P
x + (-6) = -1
solve for
x = +5</span>

5 0

3 years ago

Read 2 more answers

Is car exhaust flammable?

zavuch27 [327]

Yes, I mean it is the exhaust

5 0

3 years ago

Show that the ideal c/a ratio (height of unit cell divided by edge length) for the HCP unit cell is 1.633. (You may wish to refe

luda_lava [24]

Answer:

The structure in the first image file attached below shows the arrangement of atoms in hexagonal close packing

we need to show that the ratio between the height of the unit cell divided by its edge length is 1.633

In the structure, the two atoms are shown apart. But in fact the two atoms are touching. Therefore, the edge length is the sum of the radius of two atoms. If we assume r as radius then the expression for edge length will be as follows.

a = 2r ..............(1)

other attached files show additional solutions in steps

7 0

3 years ago

In NMR if a chemical shift(δ) is 211.5 ppm from the tetramethylsilane (TMS) standard and the spectrometer frequency is 556 MHz,

Vika [28.1K]

Answer:

The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

$\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}$

$\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}$

$Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz$

<u>Therefore, the signal is at 11759 Hz from the TMS.</u>

6 0

4 years ago