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3 years ago

What countries will see the solar eclipse?

1 answer:

4 0

You are probably still talking about the eclipse that happened last week on August 21. Totality was seen only in the United States. Parts of Canada and Mexico saw partial phases.

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What is the tangent function used to find and what are its units

Verizon [17]

The functions of angles are used to find unknown lengths or angles that can't be measured, in terms of known quantities. The trig functions of angles are ratios of lengths, so they're bare naked numbers without units.

7 0

3 years ago

A rubber ball is dropped and bounces vertically from a horizontal concrete floor. If the ball has a speed of 3 m/s just before s

Ganezh [65]

Answer:

F=12.5N

Explanation:

Net force = rate of change of momentum

$F = m*a$

so find the change of momentum P

Pdown

$P=m*v_1=0.42kg*3m/s$

Pup

$P=m*v_1=0.42kg*6.5m/s$

dP = change in P

$dP= 0.42kg (3- -.6.5)m/s =3.99 kg m/s$

$dT = 0.32 s$

$F = \frac{dP}{dt}=\frac{3.99Kg*m/s}{0.32s} =12.468 N$

6 0

4 years ago

Draw sodium formate by placing atoms on the grid and connecting them with bonds. Include all lone-pair electrons.

nikklg [1K]

Answer:

Sodium formate is the sodium salt of formic acid which is given as HCOONa.

Explanation:

The basic structure of Sodium formate consists of following bonds:

The main Ionic bond between the $HCOO^-$ radical and $Na^+$ .
The sigma covalent bonds between atom of H, atom of C and both atoms of O.
The pi bond between atom of C and one atom of O.

The structure along with lone pairs is given as attached

7 0

3 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

2 years ago

g Using the absorbance vs wavelength spectrum where the 1st peak is slightly less intense than the 2nd peak, determine the wavel

Andre45 [30]

Answer:

The wavelength range is always used to know the probable material present

Explanation:

The wavelength variation from with the concentration shows the type of material in as much the Spectrometer is well initialized before running the sample. The peaks interval may have effect on band gap.

5 0

4 years ago