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4 years ago

Which of the following is group 1? a. alkali metals b. alkaline earth metals c. halogens d. transition metals

Physics

2 answers:

scoundrel [369]4 years ago

6 0

I might be D i think try D

Elanso [62]4 years ago

4 0

The answer is A. Alkali metals

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My name is scooter hey fat boy you waa fight

zavuch27 [327]

Answer:

HEHE

YES

Explanation:

6 0

3 years ago

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The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb

balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = $F_1=5.6 lb$

Area of the piston = $A_1=0.6 inches^2$

Force applied on the brakes = $F_2$

Area of the brakes = $A_2=1.8 inches^2$

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

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

$F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb$

16.8 lb is the force on the brake pad of one wheel.

5 0

4 years ago

<img src="https://tex.z-dn.net/?f=2.25%20%5Ctimes%2030" id="TexFormula1" title="2.25 \times 30" alt="2.25 \times 30" align="absm

Paraphin [41]

67.5 is the answer i believe!!

7 0

3 years ago

Read 2 more answers

Use a common denominator to find-2/3 + -4/5

Juliette [100K]

Answer:

-22/15

Explanation:

the least common denominator is 15 so first you multiply -2/3 by 5 in both the numerator and denominator making it -10/15

Then you do the same to -4/5 except you multiply the numerator and denominator by 3 giving you -12/15

If you add -10/15+ -12/15 you get -22/15

6 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