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3 years ago

14

The device is a burner from an electric stove. It is used to transfer heat to a pot by A. Conduction B. Convection C. Radiation

D. Friction

2 answers:

OlgaM077 [116]3 years ago

4 0

<u>Answer:</u>

A device is a burner from an electric stove. Then the <em>conduction heat</em> is transferred to a pot.

Conduction heat is a method of heat transfer in which the the transfer is takes place by direct contact between objects. From the question , it is clear that, the burner from electric stove transfers energy to a bottom of the pot sitting on it.

Irina-Kira [14]3 years ago

3 0

I’m pretty positive that it’s A. Conduction. Just refer to the definition of it.

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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

Why does a light bike have more kinetic energy

FrozenT [24]

Because it doesn't use energy it uses mechanical and kinetic

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3 years ago

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball

Fudgin [204]

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

$V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\ mm^3$

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

$n = 520000\times 0.523\ grains\\\\n = 271960\ grains$

8 0

3 years ago

Please, I need help with this question.

Jet001 [13]

The x and y components of the velocity vector is 17.32 m/s and 10 m/s respectively.

<h3>What is the x - component of the velocity?</h3>

The x-component of the ball's velocity is the velocity of the ball in the horizontal direction or x-axis.

The velocity of the ball in x-direction is calculated as follows;

Vx = V cosθ

where;

Vx is the horizontal velocity of the ball
V is the speed of the ball
θ is the angle of inclination of the speed

Vx = (20 m/s) x (cos 30)

Vx = 17.32 m/s

The velocity of the ball in y-direction is calculated as follows;

Vy = V sinθ

where;

Vy is the vertical velocity of the ball
V is the speed of the ball
θ is the angle of inclination of the speed

Vy = 20 m/s x sin(30)

Vy = 10 m/s

Learn more about x and y components of velocity here: brainly.com/question/18090230

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1 year ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago