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3 years ago

An object that has negative acceleration is definitely doing what?

1 answer:

3 0

If an object is consist of a negative acceleration or composed of it, it is likely engaging of having to accelerate in a certain direction in which the direction is heading towards the opposite of the direction that is stated and that is likely to be a positive direction.

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An electron is moving directly toward you in a horizontal path when it suddenly enters a uniform magnetic field that is either v

blagie [28]

Answer:

To your left

Explanation:

The direction of the force exerted on charged particle due to a magnetic field is given by the right-hand-rule, where:

- The index finger indicates the direction of motion of the electron

- the middle finger gives the direction of the magnetic field

- the thumb gives the direction of the force if the particle is positively charged - otherwise, the direction is reversed

in this case, we have an electron (so, a negatively charged particle):

- The direction of motion (index finger) is horizontal, toward you

- The electron begins to curve upward as it enters the field, so this means that the force exerted on the electrons is upward --> the thumb must point downward (because the electron is negatively charged)

- The index finger gives us the direction of the magnetic field: therefore, to your left.

3 0

3 years ago

A golf ball is hit with an initial velocity of 20m/s at an angle of 30° to the horizontal. calculate;

svp [43]

Answer:

tcucugxojfjgfojcigxuogoudyifodtufukdutfuocyjxuogu

3 0

2 years ago

Which statement is a characteristic of a concave lens?

Trava [24]

D. it is thickest in the middle

3 0

3 years ago

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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:

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

3 0

3 years ago

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A force of 6N and another of 8N can be applied together to produce the effect of how much newtons?

const2013 [10]

12N because you are just adding those two up on the same side

5 0

3 years ago

Read 2 more answers