Why does of beam of light ben when it passes into a new medium?

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child appl

lozanna [386]

Answer:

The value is $KE = 259.6 \ J$

Explanation:

From the question we are told that

The weight of the horizontal solid disk is $W = 805 \ N$

The radius of the horizontal solid disk is $r = 1.58 \ m$

The force applied by the child is $F = 49.5 \ N$

The time considered is $t = 2.95 \ s$

Generally the mass of the horizontal solid disk is mathematically represented as

$m_h = \frac{W}{ g}$

=> $m_h = \frac{805}{ 9.8 }$

=> $m_h = 82.14 \ N$

Generally the moment of inertia of the horizontal solid disk is mathematically represented as

$I = \frac{1}{2} * m * r^ 2$

=> $I = \frac{1}{2} * 82.14 * 1.58^ 2$

=> $I = 102.5 \ kg \cdot m^2$

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

$T = I * \alpha = F * r$

=> $\alpha = \frac{ F * r }{ I }$

=> $\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }$

=> $\alpha = 0.7628$

Gnerally from kinematic equation we have that

$w = w_o + \alpha t$

Here $w_o$ is the initial angular velocity velocity of the horizontal solid disk which is $w_o = 0\ rad/s$

So

$w = 0 + 0.7628 * 2.95$

=> $w = 2.2503 \ rad/s$

Generally the kinetic energy is mathematically represented as

$KE = \frac{1}{2} * I * w^2$

=> $KE = \frac{1}{2} * 102.53 * 2.2503 ^2$

=> $KE = 259.6 \ J$

8 0

2 years ago

The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters

skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C (3)

ΔT = 5.39x10⁹ / 4200 * 75000

Hope this helps

5 0

3 years ago

A vertical spring has a mass hanging from it, which is displaced from the equilibrium position and begins to oscillate. At what

Orlov [11]

Answer:

the object has least potential energy at mean position of the SHM

Explanation:

If a block is connected with a spring and there is no resistive force on the system

In this case the total energy of the system is always conserved and it will change from one form to another form

So here we will say that

Kinetic energy + Potential energy = Total Mechanical energy

As we can say that total energy is conserved so here we have least potential energy when the system has maximum kinetic energy

So here we also know that at mean position of the SHM the system has maximum speed and hence maximum kinetic energy.

So the object has least potential energy at mean position of the SHM

5 0

3 years ago

What is the velocity of an object that has been in free fall for 1.5s?

Crank

Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

8 0

3 years ago

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

3 years ago