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Natali [406]
4 years ago
6

2. With regard to the pH scale, a solution with a pH

Physics
1 answer:
Alexandra [31]4 years ago
8 0

Answer:

ph of 10 b your welcome

Explanation:

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Several years from now you have graduated with an engineering/physics degree from OSU and have been hired by a nanoengineering f
spayn [35]

Complete Question in order

See the first image attached

Answer and Explanation:

The three images attached discussed the solution to this question

second Image                                                                                      

Third Image                                                                                          

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5 0
3 years ago
Supercells are updrafts that may cause which of the following?
ale4655 [162]

just for anyone looking for the answer i just took the test and the answer is tornadoes

6 0
3 years ago
Torque is the rotational quantity most closely related to the linear quantity
jek_recluse [69]

Answer:

The definition of torque states that one or both of the angular velocity or the moment of inertia of an object are changing.

Velocity and momentum

Explanation:

4 0
3 years ago
What is letaral inversion?​
Sphinxa [80]

Answer:

Lateral inversion is a phenomenon in which left appears to be right and vice versa. It is due to direction that light follows when it strikes a reflecting surface, generally a mirror.

8 0
3 years ago
Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 4.50 m above
Virty [35]

8.98×10^6\:\text{m}

Explanation:

First we need to find the acceleration due to gravity on the planet. The wrench took 0.809 s to fall from a height of 4.50 m so we can use the equation

y = -\frac{1}{2}gt^2

Solving for g, we get

g = -\dfrac{2y}{t^2} = -\dfrac{2(-4.50\:\text{m})}{(0.809\:\text{s})} = 13.8\:\text{m/s}^2

Recall that the acceleration due to gravity on a planet's surface can be written as

g = G\dfrac{M_p}{R_p^2}

We can express the mass of the planet M_p in terms of its density \rho as follows:

M_p = \rho \left(\dfrac{4\pi}{3}R_p^3\right) = \dfrac{4\pi}{3}\rho R_p^3

The expression for g then becomes

g = \dfrac{G}{R_p^2} \left(\dfrac{4\pi}{3}\rho R_p^3\right) = \dfrac{4\pi G}{3}\rho R_p

Solving for R_p, we get

R_p = \dfrac{3g}{4\pi G\rho}

\:\:\:\:\:\:\:= \left[\dfrac{3(13.8\:\text{m/s}^2)}{4\pi (6.674×10^{-11}\:\text{Nm}^2\text{/kg}^2)(5500\:\text{kg/m}^3)}\right]

\:\:\:\:\:\:\:= 8.98×10^6\:\text{m}

3 0
3 years ago
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