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Dmitry_Shevchenko [17]
3 years ago
9

What are concave and convex mirrors?

Physics
1 answer:
ycow [4]3 years ago
4 0
They are two different mirrors but are very similar. concave are like those round kinds that make stuff look big and convex mirrors are those that look like cameras that pop out.
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A student wanted to investigate whether or not the temperature of the water would affect the solubility of sugar. She had three
erastova [34]

ind variable ... temp

dep variable .... amount of sugar dissolved

5 0
3 years ago
When a driver presses the brake pedal, his car can stop with an acceleration of 5.4 m/s2. How far will the car travel while comi
notsponge [240]
Given:\\a=5.4 \frac{m}{s^2}  \\v_0= 25\frac{m}{s} \\\\Find:\\s=?\\\\Solution:\\\\s=v_0t- \frac{at^2}{2} \\\\a= \frac{\Delta v}{t} \\\\v_e=0\Rightarrow \Delta v=v_0\\\\t= \frac{v_0}{a} \\\\s= \frac{v_0^2}{a} - \frac{v_0^2}{2a} =\frac{v_0^2}{2a} \\\\s= \frac{ (25\frac{m}{s})2 }{2\cdot 5.4 \frac{m}{s^2} } \approx 57.9m\\\\C\rightarrow Correct\;answer
3 0
3 years ago
The first charge is pulling on the second. Is the second pulling on the first? Explain your reasoning
Alex73 [517]

Answer:

law of action and reaction.

Explanation:

In Newton's three laws it is established that forces act in pairs, if one body interacts with another the second interacts with the first, this is the so-called law of action and reaction.

In this case, when the first load pulls on the second, the second pulls on the first, the two forces are not canceled because each one is applied to a different body.

Therefore the magnitude of the forces is the same, but the direction is opposite and each one is applied in a body

7 0
3 years ago
Hydrogen (H-1), deuterium (H-2), and tritium (H-3) are the three isotopes of hydrogen. What are the values of a, b, and c in the
asambeis [7]

I believe it is a=1 b=2 c=3

plato students

6 0
3 years ago
Read 2 more answers
You overindulged on a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide t
iragen [17]

Answer:

17086

Explanation:

We are given that

Weight=5 kg

Distance of elbow from the weight=d=35 cm=\frac{35}{100}=0.35 m

1 m=100 cm

g=9.8m/s^2

Work done  each time on the weight=mgh=5\times 9.8\times 0.35=17.15 J

20%  means 0.20 of food energy=Work done each time on weight

Total work done in each time=\frac{17.15}{0.2}=85.75J

1 food calorie=4186 J

350 food calories=350\times 4186 J

Number of lift needed=\frac{350\times 4186}{85.75}=17086

Hence, you must do 17086 arm raises to work off 350 food calories.

6 0
4 years ago
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