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3 years ago

What is another example of convection

2 answers:

5 0

Radiator. A radiator puts warm air out at the top and draws in cooler air at the bottom. steaming cup of hot tea - The steam you see when drinking a cup of hot tea indicates that heat is being transferred into the air. ice melting/Ice melts because heat moves to the ice from the air

ad-work [718]3 years ago

4 0

Answer:

boiling water - When water heats, the heat from the burner is transferred to the container, heating the water at the bottom. This hot water rises, creating a circular motion, as colder water descends to replace it.

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50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui

soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

8 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

KE = Kinetic Energy and PE = Potential Energy

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

A laboratory technician was examing a blood sample from a patient.He found a rare parasite that would cause the nausea and vomit

stira [4]

Infectious Disease Specialist?

7 0

3 years ago

Read 2 more answers

A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

5 0

3 years ago

What is the period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds

olga2289 [7]

The correct answer for this question is this one: C) 2.5s. The period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds is that 2.5 s

Here are the following choices:
A) 0.25s
B) 0.40s
C) 2.5s
D) 4.0s

7 0

3 years ago