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Darya [45]
2 years ago
9

Which of the following is an example of convection

Physics
2 answers:
Fantom [35]2 years ago
6 0

Answer:

Hot air rising and cooler air falling

Explanation:

hot air rising and cooler air falling

These are called "convection currents", the currents being air currents which carry heat because they are hot air - rises - and cold air - sinks. Also may explain coastal land and sea breezes. As "thermals" convection currents may help aeroplane gliders and birds to fly by "hitching a ride on the nearest convection current" as it were

natta225 [31]2 years ago
4 0

Answer:

Boiling water - The heat passes from the burner into the pot, heating the water at the bottom. Then, this hot water rises and cooler water moves down to replace it, causing a circular motion.

Radiator - Puts warm air out at the top and draws in cooler air at the bottom.

Steaming cup of hot tea - The steam is showing heat being transfered into the air.

Ice melting - Heat moves to the ice from the air. This causes the melting from a solid to liquid.

Hot air balloon - A heater inside the balloon heats the air and so the air moves upward. This causes the balloon to rise because the hot air gets trapped inside. When the pilot want to descend, he releases some of the hot air and cool air takes it place, causing the balloon to lower.

Frozen material thawing - Frozen food thaws more quickly under cold running water that if it is placed in water. The action of the running water transfers heat into the food faster.

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if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a
Whitepunk [10]

Answer:

r=0.41m

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

\tau=r\times F

Due to the definition of cross product, the magnitude of the torque is given by:

\tau=rFsin\theta

Where \theta is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when sin\theta is equal to one, solving for r:

r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m

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2 years ago
A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o
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m = mass = 5 kg

v_{i} = initial velocity = 100 m/s

v_{f} = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m(v_{f} - v_{i})

30 = 5 (v_{f} - 100)

v_{f} = 106 m/s

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2 years ago
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One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the
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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
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Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

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