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Tasya [4]
3 years ago
7

When a pot of water sits on a hot stove the metal is He did and then transferred into heat to the world of student claims that c

onduction is the only type feet transfer occurring which statement explains why the students idea is incorrect
Physics
1 answer:
blagie [28]3 years ago
4 0

Answer:

Hello your question is poorly written hence I will just give the difference between conduction and convection

answer :

Conduction occurs when there is direct contact between solids and source of heat. while convection occurs when there is transfer of thermal energy due to movement of liquid or gas

Explanation:

when a pot of water is placed on a hot stove the pot is heated through conduction because the metal pot is in direct conduct with the heat from the stove ( which is the primary source of heat in this scenerio ).

while the heat transfer from the hot pot to the water is executed through convection this is because heat transfer via convection requires the movement of liquid or gas.  as the warmer liquid moves over to the top the cooler liquid goes below inside the pot that way the whole liquid becomes warm .

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A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first
Dominik [7]

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

6 0
3 years ago
4. AN OBJECT INCREASES ITS VELOCITY
Cloud [144]

Answer:

it is accelerating 14 m/s

Explanation:

7 0
3 years ago
A 1300-kg car moving on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to
slega [8]

The stiffness constant of the spring is 68,290.3 N/m

<h3> Stiffness constant of the spring</h3>

Apply the principle of conservation of energy;

U = K.E

¹/₂kx² = ¹/₂mv²

kx² = mv²

k = mv²/x²

where;

  • m is mass
  • v is speed = 60 km/h = 16.67 m/s
  • x is the distance

k = (1300 x 16.67²)/(2.3²)

k = 68,290.3 N/m

Thus, the stiffness constant of the spring is 68,290.3 N/m.

Learn more about stiffness constant here: brainly.com/question/1685393

#SPJ1

7 0
2 years ago
At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne
larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\  F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\  F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j

F = ma

∴

a ^{\to}= \dfrac{F^{\to}}{m}

a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)

a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j

At time t(sec; the partiCle velocity becomes v(t) = 3.78 v_o

The velocity of the charge after the time t(sec) is expressed by using the formula:

v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\  \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\  \\  0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\  \mathbf{t = 10.78 \ s}

8 0
3 years ago
Other than the depth and the atmospheric pressure, state one quantity that affects the
Hunter-Best [27]

Answer:

density

Explanation:

pressure within a liquid a liquid only depends on the density of the liquid,the acceleration due to gravity, and the depth within the liquid

7 0
3 years ago
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