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3 years ago

Which situation is an example of transferring heat by means of convection?

Physics

2 answers:

Nataly_w [17]3 years ago

6 0

Answer: b. the weather patterns outside

Explanation:

There are three modes of heat transfer:

1. Conduction: This type of heat transfer happens when there is direct contact between the two object.

2. Convection: This type of heat transfer happens when there is a movement of fluid (liquid or gas).

3. Radiation: This type of heat transfer happens when there is direct transfer of energy through space.

a. The sun warming the earth is through radiation as there is no direct contact.

b. The weather patterns outside are created due to the convection currents which arises due to movement of hot fluid upwards due to less density and the cold fluid downwards due to more density.

c. You toasty-warm in some footy pajamas is an example of conduction as there is direct contact.

d. You accidentally grabbing the hot pan from the oven is an example of conduction as there is direct contact.

kondaur [170]3 years ago

5 0

b. the weather patterns outside

There are three types of heat transfer or heat propagation; conduction, convection and radiation. Heat transfer is the process by which heat projects externally however, depending on the temperature and pressure. Also called the movement of heat from a low temperatured state which increases as heat progresses.
Conduction is the heat transfer by contact, immediate contact.
Convection is the transfer of heat through air and water.
Radiation is the transfer of heat regardless of the presence of atoms or particles.

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A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds

san4es73 [151]

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

$E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\$

=218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1) $E_{0,\pi }$

$K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}$

$u'=+-0.283c$

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

$u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c$

6 0

3 years ago

Give me one example of a sport that requires someone to have good reaction time?

DaniilM [7]

Answer:

Baseball

Explanation:

Baseball because the ball can come at you at high speeds and you need to react quickly.

4 0

3 years ago

an object with a mass of 70kilograms is supported at a height 8meters above the ground. what's the potential energy of the objec

ra1l [238]

Gravitational Potential Energy = mgh (m=mass; g=gravitational force(9.8N/kg); h = height)

Ug = (70kg)(9.8N/kg)(8m) = 5488J which is C.)

3 0

4 years ago

The staples inside a stapler are kept in place by a spring with a relaxed length of 0.116 m. if the spring constant is 46.0 n/m,

ankoles [38]

Answer:

U = 0.0207 J

Explanation:

We know that:

U = $\frac{1}{2}Kx^2$

where U is the potential energy, K the constant of the spring and x is the deformation.

so, the deformation is calcualted as:

x = 0.146m-0.116m

x = 0.03m

Finally, replacing the values of x and K, we get:

U = $\frac{1}{2}(46n/m)(0.03m)^2$

U = 0.0207 J

8 0

4 years ago

a paratrooper jumps out of a plane and opens her parachute shortly afterward. 10 seconds later the upward force on the parachute

Flura [38]

Answer:

Acceleration = 2.45 m/s²

Explanation:

The resultant vertical force on the paratrooper = ma = upward force - total weight of both the parachute and the

Ma = W - Fu .............. Equation 1

Where Fu = upward force on the parachute, W = combined Weight of the parachute and the paratrooper

W = Mg......................... Equation 2

Making M the subject of formula

M = W/g...................... Equation 3

Where M = combined mass of the parachute and the paratrooper, g = acceleration due to gravity.

Given: W = 800 N, g = constant = 9.8 m/s²

Substituting these values into equation 3

W = 800/9.8

W = 81.63 kg.

Also Given: Fu = 600 N, W = 800 N

Substituting these values into equation 1

81.63a = 800 - 600

81.63a = 200

Dividing both side of the equation by the coefficient of a

81.63a/81.63 = 200/81.63

a = 2.45 m/s²

Therefore acceleration = 2.45 m/s²

3 0

3 years ago