Ask question

Ask question

All categories

3 years ago

10

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

of static friction f on the block? Select one: A. f = mg sin θ B. f > mg cos θ C. f > mg sin θ D. f = mg cos θ E. f > mg

1 answer:

Vesnalui [34]3 years ago

7 0

Ya know what the day you are talking to you about it and then send me the back link you got it and I don’t want

You might be interested in

What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified

oee [108]

Answer:

hello your question is not properly arranged attached below is the arranged table and solution

answer : attached table below

Explanation:

Given data:

02 molecules size = 10^-10m

smoke particles size = 0.3 mm

cloud droplets size = 20 mm

Rain droplets size = 3 mm

Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths

Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.

and For Mie scattering the wavelength is the same as the wavelength.

3 0

3 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

$y=14.2* (0.9172535661)^{(31.8)}$ $y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}$

$y=14.2* 0.0641450581$ $y=14.2*(0.5)^{3.962518068}$

$y=0.9108598257$ $y=14.2* 0.0641450581$

$y=0.9108598257$

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0

1 year ago

Does the construction of a high building slow down the rotation of the earth

omeli [17]

No, the building's size in comparison to the earth would have no change or change so increadibly miniscule, like if you were told to spin slowly and an and was placed on top of your head

5 0

3 years ago

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t

BabaBlast [244]

Answer:

$-\frac{8\pi}{3}rad/s^2$

Explanation:

To solve this problem we need to apply the concept related to Angular Acceleration. We can find it through the equation

$\omega_f^2-\omega_i^2=2\alpha\theta$

Where for definition,

$\omega_i = \frac{\theta}{t}$

The number of revolution $(\theta)$ was given by 20 times, then

$\omega_i = \frac{20*2pi}{5}$

$\omega = 8\pi rad/s$

We know as well that the salad rotates 6 more times, therefore in angle measurements that is

$\theta = 6*2\pi rad = 12\pi rad$

The cook at the end stop to spin, then using our first equation,

$0-8\pi = 2\alpha (12\pi)$

re-arrange to solve $\alpha$ ,

$\alpha = \frac{-8\pi}{2*12\pi}$

$\alpha = -\frac{8\pi}{3}rad/s^2$

We can know find the required time,

$\omega_f-\omega_i = \alpha t$

Re-arrange to find t, and considering that $\omega_f=0$

$t= \frac{\omega_i}{\alpha}$

$t=\frac{-8\pi}{-8\pi/3}$

$t=3s$

Therefore take for the salad spinner to come to rest is 3 seconds with acceleration of $-\frac{8\pi}{3}rad/s^2$

6 0

3 years ago

Read 2 more answers

URGENT BY THE WAY!

Nastasia [14]

Answer:

So Nessa went so fast it made her pass the tile in a second. Lets take a look at this problem, It says "the" tile so we should assume that it means 1 tile. Then draw a diagram representing that tile then you should have your problem finished. Hope that helped and I'm willing to help if you have anymore questions!

Explanation:

6 0

2 years ago