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3 years ago

11

If the mass of each object is the same, 18 g, which object will have the largest density? *

1 answer:

ElenaW [278]3 years ago

6 0

All of them will have the same density

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the roque requried to turn the crank on an ice cream maker is 4.50 N.m how much work does it take to turn the crank through 300

Alexus [3.1K]

Answer:

the work required to turn the crank at the given revolutions is 8,483.4 J

Explanation:

Given;

torque required to turn the crank, T = 4.50 N.m

number of revolutions, = 300 turns

The work required to turn the crank is given as;

W = 2πT

W = 2 x 3.142 x 4.5

W = 28.278 J

1 revolution = 28.278 J

300 revlotions = ?

= 300 x 28.278 J

= 8,483.4 J

Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J

4 0

3 years ago

A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0

a_sh-v [17]

Answer:

a. $-12.7 Nm$

b. $-7.9 rad/s^2$

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

$F_1 = 90.0N\\F_2 = 125N$

Other parameters given include:

Mass of solid disk, $M = 24.3kg$

and radius of solid disk, $r = 0.364m$

a.) The formula for determining torque ( $T$ ), is $T = r * F$

Hence the net torque produced by the two forces is given as a summation of both forces:

$T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm$

b.) The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

$I_{disk} = {\frac{1}{2}}Mr^2$

where $M$ = Mass of solid disk

and $r$ = radius of solid disk

We then relate the torque and angular acceleration ( $\alpha$ ) with the formula:

$T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha = -{\frac{12.7}{1.61}} = -7.9 rad/s^2$

4 0

4 years ago

Which is the smallest unit of life?

emmasim [6.3K]

Answer:

the cell is the smallest unit

7 0

2 years ago

All of these are types of electromagnetic waves except for:

dimaraw [331]

C- Sigma Waves, This is what my teacher used to get us to remember https://www.youtube.com/watch?v=bjOGNVH3D4Y

3 0

3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

3 years ago