The amount of matter I’m an object is know as

A river current has a velocity of 5 km/h relative to the shore. A boat moves in the same direction as the current at 4 km/h rela

AleksAgata [21]

<span>Add the river’s velocity to the boat’s velocity.</span>

7 0

3 years ago

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A cylinder-piston system contains an ideal gas at a pressure of 1.5 105 pa.

Sedbober [7]

The change in the internal energy of the ideal gas is determined as -28 J.

The work done on the ideal gas is calculated as follows;

w = -PΔV

w = -1.5 x 10⁵(0.0006 - 0.0002)

w = -60 J

<h3>Change in the internal energy of the gas</h3>

ΔU = w + q

ΔU = -60J + 32 J

ΔU = -28 J

Thus, the change in the internal energy of the ideal gas is determined as -28 J.

Learn more about internal energy here: brainly.com/question/23876012

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5 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

3 years ago

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The lubrication of bone joints is a subject of ongoing medical research. Two bones connected at a joint do not touch. The bones

maks197457 [2]

The question is incomplete. The complete question is :

To measure the effective coefficient of friction in a bone joint, a healthy joint (and its immediate surroundings) can be removed from a fresh cadaver. The joint is inverted, and a weight is used to apply a downward force F⃗ d on the head of the femur into the hip socket. Then, a horizontal force F⃗ h is applied and increased in magnitude until the femur head rotates clockwise in the socket. The joint is mounted in such a way that F⃗ h will cause clockwise rotation, not straight-line motion to the right. The friction force will point in a direction to oppose this rotation.

Draw vectors indicating the normal force n⃗ (magnitude and direction) and the frictional force f⃗ f (direction only) acting on the femur head at point A.

Assume that the weight of the femur is negligible compared to the applied downward force.

Draw the vectors starting at the black dot. The location, orientation and relative length of the vectors will be graded

Solution :

The normal force represented by N is equal to the downward force, $F_d$ which is equal in magnitude but it is opposite in direction.

Also the frictional force acts always to oppose the motion because the bone starts moving in a clockwise direction. The frictional force that will be applied to the right direction so that the movement or the rotation at A is opposed.

5 0

3 years ago

The amount of energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin is called its

son4ous [18]

Specific heat. The definition of specific heat is the amount of energy required to raise the temperature of 1g of a substance by 1K or 1°C.

8 0

4 years ago

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