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3 years ago

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A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

ork on the hoop. What is the new speed of the center of mass of the hoop (in m/s)? Round your answer to the nearest whole number.

1 answer:

Scilla [17]3 years ago

3 0

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

now we will have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}$

$KE = mv^2$

now by work energy theorem we can say

$W = KE_f - KE_i$

$842 J = mv_f^2 - mv_i^2$

$842 = 3(v_f^2) - 3\times 11^2$

now solve for final speed

$v_f = 20 m/s$

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Which of the following statements are true? Select all that apply.

beks73 [17]

Answer:

The following options are true based on the properties of electric field;

a) Electric field lines near positive point charges radiate outward.

b) The electric force acting on a point charge is proportional to the magnitude of the point charge.

d) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.

Explanation:

From option b) From coulomb's law F = Kq1q2r/r2

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4 years ago

A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet

astraxan [27]

Answer:

Acceleration = 2.35 m/ $s^{2}$

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = $\frac{mg}{2}$ = 0.5 mg

Friction along incline = umg cos30° = mg $0.3\times \frac{\sqrt{3}}{2}$

Friction along incline = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = $\frac{net force}{mass}$ = 0.24 g = 2.35 m/ $s^{2}$

The height of incline = 8 m

Length of the inclined edge = 16 m

$v^{2}=u^{2}+2as$

$v^{2}= 2\times 0.24 \times 9.8\times 16$

v= 8.67 m/s

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4 years ago

A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density

Digiron [165]

Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude $mg$ , where m = 5.32 kg is the mass of the block and $g=9.8 m/s^2$ is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude $kx$ , where $k=179N/m$ is the spring constant and $x$ is the elongation of the spring

- The buoyant force, upward, of magnitude $\rho V g$ , where $\rho=1000 kg/m^3$ is the water density and V the volume of the block

Since the block is in equilibrium, the net force is zero, so we can write

$mg+kx-\rho V g=0$ (1)

We have to find the volume of the block first. We have:

m = 5.32 kg (mass)

$\rho_w = 622 kg/m^3$ (wood density)

So, the volume is

$V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3$

So now we can re-arrange eq.(1) to find the elongation of the spring, x:

$x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m$

So, the spring is compressed by 0.18 m.

7 0

3 years ago

A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

Blababa [14]

Answer:

(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

$mg sin \theta - \mu N = ma$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, $\theta=39^{\circ}$ being the angle of the incline

$\mu N$ is the frictional force, with $\mu = 0.6$ being the coefficient of kinetic friction, N being the normal reaction of the plane

a is the acceleration

The equation of the force along the direction perpendicular to the slope is

$N-mg cos \theta =0$

where $mg cos \theta$ is the component of the weight in the direction perpendicular to the slope. Solving for N,

$N=mg cos \theta$

Substituting into (1), solving for a, we find the acceleration:

$a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2$

6 0

4 years ago

Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, w

olga nikolaevna [1]

To solve this problem, apply the concepts related to the calculation of the work performed according to the temperature change (in an ideal Carnot cycle), for which you have to:

$W = \int\limit_{T_c}^{T_H} C (1-\frac{T_H}{T})$

Where,

C = Heat capacity of the Brick

$T_C$ = Cold Temperature

$T_H$ = Hot Temperature

Integrating,

$W = C (T_H-T_C)- T_H C ln (\frac{T_H}{T_C})$

Our values are given as

$T_H= 300K$

$T_C = 150K$

Replacing,

$W = (1) (300-150)-300(1)ln(2)$

$W = 150-300ln2$

$W = -57.94kJ \approx 58kJ$

Therefore the work perfomed by this ideal carnot engine is 58kJ

5 0

4 years ago