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2 years ago

Which type of friction prevents a pile of rocks from falling apart?

Physics

2 answers:

andrew11 [14]2 years ago

3 0

Static friction because the rocks arent moving are they?

irina [24]2 years ago

3 0

It is NOT Cohesive Friction, I just took the quiz. I believe it is the other answer.

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Which of the following would be a valid method to increase the buoyant force acting on an object?

shepuryov [24]

I think it’s B hope this helps.

4 0

2 years ago

Read 2 more answers

Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment

Drupady [299]

Answer:

$u_K=0.862$

Explanation:

The force of friction between the quails feet and the ground is:

$F=m*a$

$F_K=m*a$

$F_K=u_k*m*g$

$u_K*m*g=m*a_c$

$u_K*g=a$

$u_K=\frac{a_c}{g}$

$a_c=\frac{v^2}{r}$

So the coefficient of static is solve

$u_K=\frac{\frac{v^2}{r}}{g}$

$u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}$

$u_K=0.862$

4 0

2 years ago

Which statement is true? Which statement is true? An orbital that penetrates into the region occupied by core electrons is more

masya89 [10]

Hey there!:

Here the Statement - D is correct.

Because Orbitals containing the core electrons are more attracted towards nuclear charge and hence less shilded from nuclear charge than an orbital that doesn't penetrate. Also due to more attraction between the orbital containing core electron and nucleus, it will have less energy.

Hope this helps!

6 0

2 years ago

Compare and contrast the inner and outer planets. (Must be in complete sentences) (At least 5 sentences).

Karolina [17]

To contrast inner and outer planets we will start with the climate of the planets and then move on to there lighting. To start the planets closet to the sun, mercury, venus, earth and mars, are all hot compared to the further one, jupiter, saturn, uranus, neptune. This distance also makes the farthe away planets darker than the ones closer. Now to compare all the planets vary from either gass or solid, rocky or icy. All of them spin around the sun and all have objects spinning around them, moons.

8 0

3 years ago

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)

8 0

2 years ago