Sound travels 2146 m through a material in 1.4 seconds. What is the material?

If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus

anastassius [24]

Answer:

$E=477.92\ W.m^{-2}$

Explanation:

Given that:

Absolute temperature of the body, $T=273+30=303\ K$

emissivity of the body, $\epsilon=1$

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

$E=\epsilon. \sigma.T^4$

where:

$\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}$ (Stefan Boltzmann constant)

Now putting the respective values:

$E=1\times 5.67\times 10^{-8}\times 303^4$

$E=477.92\ W.m^{-2}$

5 0

1 year ago

In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move

lara31 [8.8K]

Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.

The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.

8 0

1 year ago

An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

arsen [322]

Kinetic energy is calculated through the equation,

KE = 0.5mv²

At initial conditions,

m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

V = 0.2977 m/s

The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J.

7 0

1 year ago

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0

1 year ago

During a family trip to Laura's grandmother's house, the family cast traveled a distance of of 8 miles in 24 minutes. During the

Vanyuwa [196]

The correct answer is B

4 0

1 year ago