Which type of force can act through empty space?

A hoop and a disk with uniform mass distribution have the same radius but the total masses are not known. Can they both roll dow

ser-zykov [4K]

Answer:

Explanation:

radius of hoop and the radius of disk is same = R

Let the mass of hoop is M and the mass of disk is M'.

As they reach the bottom of teh surface in same time so they travel equal distance thus, they have same acceleration.

The acceleration is given by

$a=\frac{gSin\theta }{1+\frac{I}{MR^{2}}}$

As the acceleration is same so that the moment of inertia is also same.

Moment of inertia of disk = moment of inertia of hoop

1/2 x mass of disk x R² = mass of hoop x R²

So, mass of disk = 2 x mass of hoop

Option (c) is correct.

5 0

3 years ago

Three monkeys A, B, and C weighing 20, 26, and 25 lb, respectively, are climbing up and down the rope suspended from D. At the i

Marina86 [1]

Answer: $2235.2lb-ft/s^2$

Explanation:

Given

Mass of monkey A=20lb

Mass of monkey B=26lb

Mass of monkey C=25lb

acceleration of monkey A= $4.2ft/s^2$

acceleration of monkey B=0

acceleration of monkey C= $-5.4ft/s^2$

Force Due to monkey A $\left ( F_A\right )=20\times 4.2 =84lb-ft/s^2\left ( downwards\right )$

Force Due to monkey A $\left ( F_B\right )=26\times 0 =0lb-ft/s^2$

Force Due to monkey A $\left ( F_C\right )=25\times 5.4 =135 lb-ft/s^2\left ( upward\right )$

In addition to it Weights of monkeys will be acting downwards therefore net Downwards force is balanced by tension

T= $\left ( 20+26+25\right )32.2+84-135=2235.2 lb-ft/s^2$

5 0

3 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

3 years ago

A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional d

Law Incorporation [45]

Answer:

Third displacement = 2.81 m which is 61.70° north of east.

Explanation:

Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

She sails 2 km east, displacement = 2 i

Then 3.50 km southeast, means 3.5 km 315⁰ to + ve X axis

Displacement = 3.5 cos 315 i + 3.5 sin 315 = 2.47 i - 2.47 j

Let third displacement be x i + y j

We have final displacement = 5.80 km east = 5.80 i

From summation we have total displacement = 2 i + 2.47 i - 2.47 j + x i + y j

= (4.47+x) i + (y - 2.47) j

Comparing both , we have 4.47+x = 5.80

x = 1.33

y-2.47=0

y = 2.47 j

So third displacement = 1.33 i + 2.47 j

Magnitude of third displacement = $\sqrt{1.33^2+2.47^2} =2.81m$

θ = tan⁻¹(2.47/1.33) = 61.70°

So third displacement = 2.81 m which is 61.70° north of east.

7 0

3 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Which way(s) can increase the c

Charra [1.4K]

Explanation:

When an object moves in a circular path, due to the change in its velocity, the object possess centripetal acceleration. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

Where

v is the speed of the object

R is the radius of the circle path

Option (1) : Keeping the speed fixed and decreasing the radius by a factor of 9

$a=\dfrac{v^2}{R/9}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration of the ball by a factor of 9.

Option (2) : Keeping the radius fixed and increasing the speed by a factor of 3

$a=\dfrac{(3v)^2}{R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

Option (3) : Decreasing both the radius and the speed by a factor of 9.

$a=\dfrac{(v/9)^2}{R/9}$

$a=\dfrac{(v)^2}{9R}$

The centripetal acceleration decreases by a factor of 9.

Option (4) : Keeping the radius fixed and increasing the speed by a factor of 9

$a=\dfrac{(3v)^2}{R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

Option (5) : Increasing both the radius and the speed by a factor of 9

$a=\dfrac{(9v)^2}{9R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

Option (6) : Keeping the speed fixed and increasing the radius by a factor of 9

$a=\dfrac{(v)^2}{9R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

So, as the radius of the circle decreases, its centripetal acceleration increase. Also, if the speed of the object increases, its centripetal acceleration increase. Hence, this is the required solution.

4 0

3 years ago