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2 years ago

Please help me-. it’s part of my major grade.

Physics

2 answers:

stich3 [128]2 years ago

7 0

It will continue to not move- if you leave an object alone it’s not going to move unless something happens to it - I.e an outside force.

Let me know if you have any questions

Serga [27]2 years ago

6 0

Answer:

Explanation:

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If a car is moving at a constant velocity of 10 m/s, what is its acceleration?

marin [14]

Answer:

If a car is moving at a constant velocity of 10 m/s, there will be no change in the velocity per time.

Acceleration is the rate of change of velocity

Acceleration=(Final velocity-Initial velocity)/Time

Acceleration=(10m/s-10m/s)/Time= 0 m/s²

<h3>★ 0 m/s² is the right answer. </h3>

4 0

2 years ago

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:

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

3 0

2 years ago

Read 2 more answers

Technician A says that hill assist and hill descent controls are added features to some electronic stability control systems. Te

Pani-rosa [81]

Answer:

Both technicians A and B

Explanation:

Both trailer sway control, hill assist and hill descent controls are additional featires that enhance stability of electronics and their control systems. Majorly, these features track and reduce skidding in electronics, therfore, enhancing electronic system stability. During the process, these newly added features help to automatically apply brakes and direct the sytem where the controller wants to take it.

8 0

2 years ago

Why is it important for you to understand the basics of organ systems?

marissa [1.9K]

This question can have ALOT of answers but ill leave you with these summed up points and you can take what you need from it they are get right to the point! Sorry if they long paragraphs scare you lol

*You want to provide patients the best care possible. Most often your patients will have a disease. Diseases result when there is something abnormal in the anatomy and physiology of a structure. With a car, you can’t understand how to fix an engine if you don’t know how it works. The same is true with your patients. You can’t really understand how to treat them or why the treatment works, if you don’t understand how the effected body system normally functions.

*Patients will want to understand their diseases. In order to help them understand what is going wrong, you have to first understand how a particular organ is supposed to work. In addition, you will need to be able to explain these things to patients in a way that they can understand. If you don’t understand it well, you won’t be able to explain it. Your patient’s confidence in your ability will be at least partially determined by your ability to discuss what you are doing and why you are doing it. You will need to look up information if you are not sure.

*Organ systems are so interconnected that a disease in one system may result in a symptom in another system. Without seeing the normal interconnectedness, you cannot fully understand the disease.

*Success in an allied health field requires at least three things. First, you must have the personality to be able to support and help patients. Secondly, you must have the scientific and technical knowledge necessary to make the correct decisions regarding patient care. Thirdly, you must have the clinical skills necessary to implement this kno

5 0

2 years ago

Starting at (0,0) an object travels 36 meters north and then it covers 20 meters east. What is

Svetradugi [14.3K]

Answer:

Explanation:

Using the pythagoras theorem, the displacement is expressed as;

d² = x²+y²

y = 36m (north)

x = 20m east

Substitute;

d² = 36²+20²

d² = 1296+400

d² = 1696

d = √1696

d = 41.18m

For the direction;

theta = tan^-1(y/x)

theta = tan^-1(36/20)

theta = tan^-1(1.8)

theta = 60.95°

Hence the magnitude is 41.18m and the direction is 60.95°

8 0

3 years ago