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3 years ago

A salt shaker sits 0.102 m from

Physics

1 answer:

Ivahew [28]3 years ago

6 0

The maximum speed is 0.55 m/s

Explanation:

For an object in uniform circular motion, the force of friction between the object and the ground provides the centripetal force required to keep the body in motion. Therefore we can write:

$\mu_s mg = m\frac{v^2}{r}$

where the term on the left is the frictional force and the term on the right is the centripetal force, and where

$\mu_s$ is the coefficient of static friction

m is the mass of the body

g is the gravitational acceleration

v is the speed of the body

r is the radius of the circular path

In this problem, we have:

$\mu_s = 0.307$

r = 0.102 m

$g=9.8 m/s^2$

Substituting and re-arranging, we find the maximum speed v at which the salt shaker can rotate:

$v=\sqrt{\mu gr}=\sqrt{(0.307)(9.8)(0.102)}=0.55 m/s$

Learn more about circular motion:

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brainly.com/question/6372960

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You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen

Paladinen [302]

Answer:

(A)

$\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s \right )D+t_t^2v_s^2=0$

(B) D=54.71 m

Explanation:

Free Fall

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

$\displaystyle D=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2D}{g}}$

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

$\displaystyle t=\frac{D}{v_s}$

(A)

The total measured time is the sum of both times and it's given as $t_t=3.5\ seconds$

$\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}$

From this equation we'll manage to compute D

First, we isolate the square root

$\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}$

Let's square both sides

$\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}$

Multiplying by $v_s^2$

$\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2$

Rearranging and factoring

$\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}$

Now, let's put in numbers:

$g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec$

$\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0$

Computing all the coefficients:

$\displaystyle D^2-26,705.82D+1,458,056.25=0$

Solving for D, we have two possible solutions:

$D=54.71,\ D=26,651.11$

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

D=54.71 m

Let's prove our calculations by computing both times:

$\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec$

$\displaystyle t_2=\frac{54.71}{345}=0.16\ sec$

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0

3 years ago

The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re

Sever21 [200]

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

5 0

3 years ago

An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai

rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion: Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

Vf = Vi + a×t

Vf = 0 + 2.4×3.6

Vf = 2.4×3.6

Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0

2 years ago

Three metal spheres are placed as shown in the image. Sphere A weighs 5 kg, B weighs 8 kg, and C weighs 3 kg.

Anni [7]

Answer:

Explanation:

7 0

2 years ago

What mRNA sequence would result from the following DNA sequence?

Nezavi [6.7K]

Answer:

UAC CUG AGG AUC

Explanation:

The mRNA sequence from ATG GAC TCC TAG DNA sequence would be UAC CUG AGG AUC.

According to Chargaff's base pairing rule, the purine bases always pair with pyrimidine bases. Specifically, Adenine base must pair with Thymine base while Guanine base must pair with Cytosine base. In RNA, Thymine base is replaced with Uracil base.

Hence:

ATG GAC TCC TAG will pair with

UAC CUG AGG AUC

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3 years ago