Answer:
B
Explanation:
Because when you go to the doctors they say, "You need to get your x-rays done" not microwave
Hey there : !
density = 10.5 g/cm³
volume = 23.6 cm³
therefore:
D = m / V
10.5 = m / 23.6
m = 10.5 * 23.6
m = 247.8 g
hope this helps!
<span>The action of suspending someone or something or the condition of being suspended, in particular.
OR<span><span /></span>
The system of springs and shock absorbers by which a vehicle is cushioned from road conditions.</span>
Answer:
Explanation:
<em>2. A 10 kg bowling ball would require what force to accelerate down an alleyway at a rate of 3m/s² ?</em>
Notice that I completed the question with the garbled and missing values:
<u>Data:</u>
<u />
<u>Physical principles:</u>
- Newton's second law:

<u>Solution:</u>

<em></em>
<em>3. Salty has a car that accelerates at 5 m/s². If the car has a mass of 1000 kg, how much force does the car produce?</em>
Notice that I arranged the typos.
<u />
<u>Data:</u>
<u>Physical principles:</u>
- Newton's second law:

<u>Solution:</u>

<em>4. What is the mass of a falling rock if it produces a force of 147 N?</em>
<u>Data:</u>
<u>Physical principles:</u>
- neglecting air resistance ⇒ a = g: gravitational acceleration: 9.8m/s²
- Newton's second law:

<u>Solution:</u>
- Clear m from Newton's second law

- Substitute with F = 147 N and a = g = 9.8m/s², and compute

<em></em>
<em>5. What is the mass of a truck if it produces a force of 14,000 N while accelerating at a rate of 5 m/s²?</em>
<u>Data:</u>
<u>Physical principles:</u>
- Second Newton's law:

<u>Solution:</u>
- Clear m from Newton's second law

- Substitute with F = 14,000 N and a = 5m/s², and compute

The pH of a 0.0115 m aqueous formic acid solution is mathematically given as
pH=2.8424
This is further explained below.
<h3>What is the ph of a 0.0115 m aqueous formic acid solution?</h3>
Generally, the equation for the chemical equation is mathematically given as
HCOOH H^+ + HCOO


![&\left[\mathrm{H}^{+}\right]=\mathrm{C \alpha}\\\\&=0.125 \times 0.0115\\\\&=1.4375 \times 10^{-9}\\\\&P=-\log \left[H^{+}\right]\\\\&=-\log \left[1.4375 \times 10^{-3}\right]\\\\&P H=2.8424](https://tex.z-dn.net/?f=%26%5Cleft%5B%5Cmathrm%7BH%7D%5E%7B%2B%7D%5Cright%5D%3D%5Cmathrm%7BC%20%5Calpha%7D%5C%5C%5C%5C%26%3D0.125%20%5Ctimes%200.0115%5C%5C%5C%5C%26%3D1.4375%20%5Ctimes%2010%5E%7B-9%7D%5C%5C%5C%5C%26P%3D-%5Clog%20%5Cleft%5BH%5E%7B%2B%7D%5Cright%5D%5C%5C%5C%5C%26%3D-%5Clog%20%5Cleft%5B1.4375%20%5Ctimes%2010%5E%7B-3%7D%5Cright%5D%5C%5C%5C%5C%26P%20H%3D2.8424)
Read more about chemical equation
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