37. Which of the following is not used in cal-

A 2-kg bowling ball rolls at a speed of 10m/s on the ground whats its ke

kondaur [170]

Answer:

Explanation:

I wonder what level of physics you are taking? If this is a beginning course (which I'm going to assume) then the KE is

KE = 1/2 m v^2

KE = 1/2 * 2kg * (10 m/s)^2

KE = 100 Joules.

If you are in an upper level course and need to take into account the rotational motion, leave a note.

3 0

3 years ago

Help please ! USA test prep

Illusion [34]

C) 2 ohms I think don't fully trust my answer I suck at math but I think it's c

3 0

3 years ago

A rock is thrown straight up into the air. At the

Aleonysh [2.5K]

The answer is 3.

When a ball is thrown into the air, there is no friction (there's nothing rubbing against it), so the only force acting on the ball is gravity, which aims to pull the ball back down. Since this is the only force, the answer will equal the gravity acting on the ball. Since "weight" is defined as just that (the gravity acting on something), the answer is 3, the magnitude of the rock's weight (or the force of gravity).

6 0

3 years ago

A spring has a spring constant of 30000 N/m. How far must it be stretched (in meters) for its potential energy to be 47 J?

Evgen [1.6K]

The elastic potential energy in a spring is given by:
EPE = 1/2 kx²
x = √(2 x 47 / 30,000)
x = 0.056 m

5 0

3 years ago

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (a) How

Viktor [21]

Answer:

(a) $v_c=1.56\ m.s^{-1}$

(b) $KE=5.3539\ J$

(c-i) $v_h=3.12\ m.s^{-1}$ in straight rightward direction.

(c-ii) $v_l=0\ m.s^{1}$

(c-iii) $v_r=2.2062\ m.s^{-1}$ $\theta=45^{\circ}$ to the bottom of horizontal right.

(d-i) $v_h=1.56\ m.s^{-1}$ to the horizontal right.

(d-ii) $v_l=1.56\ m.s^{-1}$ horizontally left

(d-iii) $v_r=1.56\ m.s^{-1}$ moving vertically downward

Explanation:

Given:

mass of hoop, $m=2.2\ kg$

diameter of hoop, $d=1.2\ m$

angular speed of hoop, $\omega=2.6\ rad.s^{-1}$

So, time taken for 1 revolution(2π radians) of the hoop:

$T=\frac{2\pi}{2.6}\ s$

(a)

The center will move linearly in the right direction.

circumference of the hoop:

$c=\pi.d$

$c=\pi\times 1.2$

<u>Now the speed of center:</u>

$v_c=\frac{c}{T}$

$v_c=\frac{\pi\times 1.2}{(\frac{2\pi}{2.6})}$

$v_c=1.56\ m.s^{-1}$

(b)

Moment of inertia for ring about central axis:

$I=m.r^2$

where 'r' is the radius of the hoop.

$I=2.2\times 0.6^2$

$I=0.792\ kg.m^2$

∴Kinetic energy

$KE=\frac{1}{2} I.\omega^2+\frac{1}{2} m.v_c^2$

$KE=\frac{1}{2} 0.792\times 2.6^2+\frac{1}{2} 2.2\times 1.56^2$

$KE=5.3539\ J$

(c) (i)

The highest point on the hoop will have the maximum velocity.

Given by:

$v_h=\omega \times d$

$v_h=2.6\times 1.2$

$v_h=3.12\ m.s^{-1}$ in straight rightward direction.

(c) (ii)

Lowest point n the hoop will seem stationary for an observer on the ground.

$v_l=0\ m.s^{1}$

(c) (iii)

Velocity of the right-most point of the loop.

This velocity will have 2 components horizontal right and vertical down.

$v_r=\sqrt{v_c^2+v_d^2}$

here: $v_d$ is the downward component.

$v_d=r.\omega$

$v_d=1.56\ m.s^{-1}$

$\therefore v_r=\sqrt{1.56^2+1.56^2}$

$v_r=2.2062\ m.s^{-1}$

$tan\theta=\frac{1.56}{1.56}$

$\theta=45^{\circ}$ to the bottom of horizontal right.

When the observer is moving in the same direction with $v_c$ velocity:

(d) i

Then,

$v_h=\omega \times r$

$v_h=2.6\times 0.6$

$v_h=1.56\ m.s^{-1}$ to the horizontal right.

(d) ii

The bottom point of hoop will seem to move horizontally left with velocity:

$v_l=r.\omega$

$v_l=0.6\times 2.6$

$v_l=1.56\ m.s^{-1}$ horizontally left

(d) iii

Contrary to the case of stationary observer, this observer will see the extreme right point of the hoop moving vertically downward with a velocity:

$v_r=r.\omega$

$v_r=0.6\times 2.6$

$v_r=1.56\ m.s^{-1}$

5 0

3 years ago