A 34.00 kg sample of mud has a specific heat of 2,512 J/kg °C.

A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation.

dezoksy [38]

Hi there!

We can use the conservation of angular momentum to solve.

$L_i = L_f\\\\I\omega_i = I\omega_f$

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

$\text{Solid cylinder:} I = \frac{1}{2}MR^2\\\\\text{Object around center:} = MR^2$

Begin by converting rev/sec to rad sec:

$\frac{0.17rev}{s} * \frac{2\pi rad}{1 rev} = 1.068 \frac{rad}{s}$

According to the above and the given information, we can write an equation and solve for ωf.

$1.068(\frac{1}{2}(34)(1.6)^2 + (79)(1.6)^2) = \omega_f(\frac{1}{2}(34)(1.6^2) + 79(0^2))\\\\\omega_f = \boxed{6.03 \frac{rad}{sec}}$

4 0

2 years ago

A scuba diver uses his waterproof flashlight to shine a beam of light so that it strikes the surface of the water at an angle of

Mrrafil [7]

Snell's law: n1Sinα=n2Sin β where α=Incidence angle, β=angle of refraction, n1 and n2 are the indices of refraction for water and air respectively.
Therefore,
Sinα=n2/n1 Sinβ For refracted ray to be along the surface of water, β=90° and thus Sinβ = 1

Sinα=n2/n1= 1/1.33 = 0.7519 => α=sin^-1 (0.7519) = 48.75°
When light moves from a medium of higher index of refraction (such as water) to medium of lesser index of refraction (such as air), the refracted ray is bend such that α is bigger than β. This is internal refraction. At some value of α, β approaches 90°. This incidence angle is called critical incidence angle. Therefore, the current scenario is shows critical angle of incidence.

8 0

3 years ago

A weight lifter picks up a barbell and 1. lifts it chest high 2. holds it for 30 seconds 3. puts it down slowly (but does not dr

Anna11 [10]

1. lifts it chest high

The force opposing to this action is the force due to gravity. Therefore the work done is:

W1 = m g d

where m is mass of the barbell, g is gravity and d is displacement

2. holds it for 30 seconds

Work is a product of force and displacement, since there is no displacement, therefore work done is zero.

W2 = 0

3. puts it down slowly

If the barbell was dropped, then it would simply be a free fall. But since it was not, so the work done here is also equal to the weight of the barbell times displacement:

W3 = m g d

We can see that W1 = W3, and since W2 = 0, therefore the answer is:

w3 = w1 > w2

7 0

2 years ago

Which wavelength of the electromagnetic spectrum are shorter than visible light and carry more energy

geniusboy [140]

Answer:

C

Explanation:

Which wavelength of the electromagnetic spectrum are shorter than visible light and carry more energy

A micro

B radio

C ultraviolet

F infrared

The correct answer is C

The visible light wavelength is of 7 × 10^-7 m while the ultraviolet is of other

4 × 10^-8 m

Therefore, ultravolet is shorter than visible light and carry more energy

Therefore, the correct answer is option C.

7 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

For the course of upward acceleration:

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

Now the velocity of the rocket just after the fuel is finished:

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

2 years ago