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Sonbull [250]
3 years ago
14

Draw a conclusion, based on the solubility curves shown above, of which compound would have the greatest

Physics
1 answer:
Colt1911 [192]3 years ago
3 0

Answer: The answer is D. KNO3

Explanation:

The graph shows that the KN03 going straight up from the temperature sign so you reversed that so that it will make it to 90°C to 30°C

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After 2 seconds the stone will have fallen a total of 19.6m how far will the stone have fallen after 5 seconds
Reil [10]
After 5 seconds the stone will have fallen 49 m
6 0
3 years ago
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what type of simple machine is shown in the diagram?write the length and height of slope in it. Please help ASAP!!!​
Akimi4 [234]

the machine shown in the diagram is called a tramp.

the height of the tramp is 5 and the length is 12

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3 years ago
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Can you have zero displacement and nonzero average velocity? Zero displacement and nonzero velocity? Illustrate your answers on
Svetlanka [38]

a) Not possible

b) Yes, it's possible (see graph in attachment)

Explanation:

a)

The average velocity of a body is defined as the ratio between the displacement and the time elapsed:

v=\frac{\Delta x}{\Delta t}

where

\Delta x is the displacement

\Delta t is the time elapsed

In this problem, we want to have zero displacement and non-zero average velocity. From the equation above, we see that this is not possible. In fact, if the total displacement is zero,

\Delta x = 0

And therefore as a consequence,

v=0

which means that the average velocity is zero.

B)

Here we want to have zero displacement and non-zero velocity. In this case, it is possible: in fact, we are not talking  about average velocity, but we are talking about (instantaneous) velocity.

On a position-time graph, the instantaneous velocity is the slope of the graph. Look at the graph in attachment. We see that the position of the object first increases towards positive value, then it decreases (the object starts moving backward), then becomes negative, then it increases again until returning to the original position, x = 0.

In all of this, we notice that the total displacement of the object is zero:

\Delta x = 0

However, we notice that the instantaneous velocity of the object at the various instants is not zero, because the slope of the graph is not zero.

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

3 0
4 years ago
A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed
Dahasolnce [82]
<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>
6 0
3 years ago
A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
zzz [600]

Answer:

0.0257259766982 m

Explanation:

P_2 = Atmospheric pressure = 101325 Pa

d_1 = Initial diameter = 1.5 cm

d_2 = Final diameter

\rho = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m

The diameter of the bubble is 0.0257259766982 m

8 0
3 years ago
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