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RSB [31]
2 years ago
5

Instructions:Drag the tiles to the correct boxes to complete the pairs. Match each term with its definition. Tiles conductor rad

iation insulator convection conduction Pairs heat transfer involving direct contact of particles arrowBoth heat transfer in fluids arrowBoth heat transfer that doesn’t need a medium arrowBoth substance that doesn’t allow heat through arrowBoth substance that allows heat through arrowBoth
Physics
2 answers:
Vladimir [108]2 years ago
5 0

Explanation:

1. The conductor is the substance that allows the transfer of heat.

2. In the case of radiation, the transfer of heat happens without any medium.

3. The insulators do not allow the transfer of heat.

4. Convection - This process occurs in the case of fluids. The rate of convection depends on the temperature of the fluid and the temperature of the surface.

5. Conduction - It is the process of the transfer of heat when there is a collision between the existing particles.

UNO [17]2 years ago
4 0

<span>High SchoolPhysics5+3 pts</span><span>Instructions:Drag the tiles to the correct boxes to complete the pairs. Match each term with its definition. Tiles conductor radiation insulator convection conduction Pairs heat transfer involving direct contact of particles arrowBoth heat transfer in fluids arrowBoth heat transfer that doesn’t need a medium arrowBoth substance that doesn’t allow heat through arrowBoth substance that allows heat through arrowBoth

These are the answers:

</span>Conductor - <span>substance that allows heat through 
</span>Radiation - <span> heat transfer that doesn’t need a medium
</span>Insulator -  <span>substance that doesn’t allow heat through 
</span>Convection - <span>heat transfer in fluids
Conduction - </span>heat transfer involving direct contact of particles 
You might be interested in
Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f
Lunna [17]

Answer:

The final speed will be the same for the children on the shorter side and on the longer side.

Explanation:

This is because since the they are the same distance above the ground, their potential energy which is a function of mass, acceleration due to gravity and vertical height are the same.

PE = Mass × gravity × vertical height

At this point, we can deduce that the horizontal length of the slide has no effect on the potential energy. Only the vertical height does.

All this potential energy is converted to kinetic energy at the end of the slide. Since the potential energy is the same, then the kinetic energy will be the same and thus their velocity is the same.

Mathematically, consider that PE = mgh and KE = \frac{1}{2}mv^{2}

at the bottom of the slide, since energy has to be conserved, PE must be equal to KE.

mgh = \frac{1}{2}mv^{2}

final velocity of the child , v = \sqrt{2gh}

It shows the final velocity is only a function f acceleration due to gravity and height.

Thus, making their velocities equal.

8 0
1 year ago
For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?
Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph  speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity (v_t), in order to obtain the angular velocity of the wheel:

v_t=w*R\\w=\frac{v_t}{R}

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}

And now, recalling that 2\pi radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by 2\pi :

rotational frequency = \frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}

6 0
2 years ago
An ice rescue team pulls a stranded hiker off a frozen lake by throwing him a rope and pulling him horizontally across the essen
Anuta_ua [19.1K]

Answer:T=116.84 N

Explanation:

Given

Weight of hiker =1040 N

acceleration a=1.1 m/s^2

Force exerted by Rope is equal to Tension in the rope

F_{net}=T=ma_{net}

T=\frac{1040}{g}\times 1.1

T=116.84 N

8 0
2 years ago
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

 So maximum height reached = 35.15 meter.

7 0
1 year ago
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.12
den301095 [7]

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s.  What is its angular momentum? Express your answer in kilogram meters squared per second.

Answer:

0.0192 kgm^{2}/s

Explanation:

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

L = \frac{1}{{12}}m{l^2}\omega

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for \omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}

5 0
2 years ago
Read 2 more answers
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