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3 years ago

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A box that has a mass of 80 kg slides down a ramp with a 30 degree angle. The free-body diagram shows the forces acting on the b

ox. Ignoring friction and air resistance, what is the acceleration of the box, to the nearest tenth? Answer choices are 0.5 m/s2 4.9 m/s2 8.5 m/s2 9.8 m/s2

2 answers:

zimovet [89]3 years ago

8 0

The first one I believe

Nonamiya [84]3 years ago

8 0

Answer:

a= 4.9 $\frac{m}{s^{2} }$

Explanation:

Force equation

$F= m*a$

$F= W*g*sen (30)\\F= 80 kg * 9.8 \frac{m}{s^{2} } * sen (30)\\F= 80 kg * 9.8 \frac{m}{s^{2} } * 0.5\\F= 392 *\frac{kg*m}{s^{2} } = 392$ N

$a= \frac{F}{m} = \frac{392 N}{80 kg} = \frac{392 \frac{kg*m}{s^{2} } }{80 kg}\\ a= 4.9 \frac{m}{s^{2} }$

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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

Read 2 more answers

POINTS WILL BE GIVEN!! VVV SIMPLE IF YOURE A GENIUS

insens350 [35]

Answer:

9

Explanation:

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6 0

3 years ago

Read 2 more answers

A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce

Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

$x=x_0+v_0t+at^2$

Where

$x_0 =$ Initial position

$v_0 =$ Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

$v_f = v_0 + at$

Where,

$v_f = Final velocity$

For our case we have that there is neither initial position nor initial velocity, then

$x= at^2$

With our values we have $x = 401.4m, t=4.945s$ , rearranging to find a,

$a=\frac{x}{t^2}$

$a = \frac{ 401.4}{4.945^2}$

$a = 16.41m/s^2$

Therefore the final velocity would be

$v_f = v_0 + at$

$v_f = 0 + (16.41)(4.945)$

$v_f = 81.14m/s$

Therefore the final velocity is 81.14m/s

8 0

3 years ago

People who view society as a set of interrelated parts that work together to produce a stable social system are said to employ t

AURORKA [14]

The answer is d. functionalist perspective

7 0

3 years ago

Calculate the momentum of a Lion of mass 130-kg and moving at a speed of 22.3 m/s [W]

Sunny_sXe [5.5K]

Answer:

<h2>289.9 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 130 × 22.3

We have the final answer as

<h3>289.9 kg.m/s</h3>

Hope this helps you

3 0

3 years ago