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2 years ago

15

Mike understands the formula for temperature conversion, but is having difficulty imagining what various temperatures feel like.

which of these would be a good example to tell mike what 37°c feels like? a. room temperature b. a hot pan c. body temperature d. a chilly autumn evening

1 answer:

Alisiya [41]2 years ago

3 0

A good example that will help to to tell mike what 37°c feels like is body temperature.

<h3>What is room temprrature?</h3>

The average room temperature is typically around 20°C, or 68 degrees Fahrenheit.

<h3>What is body temperature?</h3>

The average body temperature is 98.6 Fahrenheit or 37 degrees Celsius.

Thus, a good example that will help to to tell mike what 37°c feels like is body temperature.

Learn more about body temperature here: brainly.com/question/5295345

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You might be interested in

2. Interpret Graphs How does temperature change with depth in Earth's<br> mantle?

fomenos

The Earth gets hotter as one travels towards the core, known as the geothermal gradient. The geothermal gradient is the amount that the Earth's temperature increases with depth. ... On average, the temperature increases by about 25°C for every kilometer of depth.

3 0

3 years ago

in which of the following collisions would you expect the kinetic energy to be conserved? a. a bullet passes through a block of

Scorpion4ik [409]

An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum and kinetic energy are both conserved.

<h3>Explain about the Elastic Collision?</h3>

A collision between two bodies in physics is referred to as an elastic collision if their combined kinetic energy stays constant. There is no net conversion of kinetic energy into other forms, such as heat, noise, or potential energy, in an ideal, fully elastic collision

An example of an elastic collision is when two balls collide at a pool table. It is an elastic collision when you throw a ball on the ground and it bounces back into your hand because there is no net change in the kinetic energy.

If there is no kinetic energy lost in the impact, the collision is said to be perfectly elastic. A collision is considered to be inelastic if any of the kinetic energy is converted to another kind of energy during the collision.

To learn more about Elastic Collision refer to:

brainly.com/question/7694106

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8 0

1 year ago

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

3 years ago

The interaction between electrical energy and magnetism has been an important

choli [55]

Answer:

Maybe

Explanation:

I say maybe because it will help them still but not quite

4 0

3 years ago

Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$

Part b)

Angular acceleration is given as

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = 176.6 N$

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

$m_1g r_1 = m_2 g r_2$

$20 g(0.5) = m_2 g(1.5)$

$10 = 1.5 m_2$

$m_2 = \frac{20}{3} kg$

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

$m_1g - T = m_1 a$

also we have

$T r_1 = I\alpha$

now we have

$m_1g = \frac{I a}{r_1^2} + m_1 a$

$a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}$

$a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}$

$a = 0.981 m/s^2$

so angular acceleration is given as

$\alpha = \frac{a}{r_1}$

$\alpha = \frac{0.981}{0.5}$

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = \frac{I\alpha}{r_1}$

$T = \frac{45 (1.96)}{0.5}$

$T = 176.6 N$

7 0

3 years ago