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Zina [86]
3 years ago
9

Visualize five horizontal sedimentary strata exposed in a cliff or canyon wall identified by consecutive numbers, 1 being the lo

west bed and 5 being the highest. Which of the following statements concerning the strata are true?
a. Bed 4 is older than bed 2.
b. Bed 3 is older than beds 2 and 4.
c. Bed 5 is the oldest.
d. Beds l and 3 are older than bed 4.
Physics
2 answers:
Grace [21]3 years ago
6 0

Answer:

Only the 4 th statement is true that is bed 1 and 3 are older than 4.

Explanation:

The 5 beds are numbered from 1 to 5 , 1 being the lowest and 5 being the topmost bed.

We are given 4 statements and we have to find out which all are true.

(<u>a)Bed 4 is older than bed 2 </u>

This is wrong because the lower beds are older than beds that are higher.

<u>(b)Bed 3 is older than beds 2 and 4</u>

This is also wrong because the 2 is older than 3

(<u>c)Bed 5 is the oldest </u>

This is wrong because bed 1 is the oldest

(<u>d)Beds 1 and 3 are older than 4</u>

This is true as lower beds are older

<u />

Vlad1618 [11]3 years ago
4 0

Answer:

bed 2 was deposited before bed 3

Explanation:

Which of the following conclusions can be made about the sedimentary layers?

bed 2 was deposited after bed 3

bed 1 is the youngest

bed 3 was deposited before bed 1

bed 2 was deposited before bed 3

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Explanation:

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Work done is a physical quantity that is defined as the force applied to move a body through a particular distance.

Work is only done when the force applied moves a body through a distance.

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The maximum work is done when the force is parallel to the distance direction.

The minimum work is done when the force is at an angle of 90° to the distance direction.

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A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig
inna [77]

Answer:

\mu_s=\frac{1}{3}\tan \theta

Explanation:

Let the minimum coefficient of static friction be \mu_s.

Given:

Mass of the cylinder = M

Radius of the cylinder = R

Length of the cylinder = L

Angle of inclination = \theta

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

\tau =I\alpha

Now, angular acceleration is given as:

\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}

Also, moment of inertia for a cylinder is given as:

I=\frac{MR^2}{2}

Therefore, the torque acting on the cylinder can be rewritten as:

\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are mg\sin \theta\ and\ f. The net force acting along the incline is given as:

F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, N=Mg\cos \theta

Plugging in N=Mg\cos \theta in equation (2), we get

F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3

Now, as per Newton's second law,

F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times  R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

\mu_s=\frac{1}{3}\tan \theta

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