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soldi70 [24.7K]

3 years ago

13

Which situation is an example of transferring heat by means of convection?

1 answer:

Hoochie [10]3 years ago

3 0

It would be B, the weather patterns outside.

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4. A family leaves from New York City and is flying to Los Angles which is 2800mi away. It takes

Blizzard [7]

The average speed of the whole travel is equal to <u>400 mph</u>.

Why?

From the statement, we know that whole travel is divided into three parts. For the first part (traveling from New York to Chicago), we have that it was 3.25 hours and the covered distance was half of the total distance (1400mi). For the second part, we have that it was 1 hour (layover time), and the covered no distance. For the third part (traveling from Chicago to Los Angeles), we have that it was 2.75 hours, and it took the other half of the total distance (1400mi).

We can calculate the average speed of the whol travel using the following formula:

$AverageSpeed=\frac{distance_{1}+distance_{2}+distance_{3}}{time_{1}+time{2}+time_{3}}$

Now, substituting and calculating, we have:

$AverageSpeed=\frac{1400mi+0mi+1400mi}{3.25h+1h+2.75h}$

$AverageSpeed=\frac{2800mi}{7h}=400mph$

Hence, we have the average speed of the whole travel is equal to 400 mph.

Have a nice day!

7 0

3 years ago

What two types of atoms make a Covalent bond

dalvyx [7]

Ionic bonds are formed between a cation (metal) and an anion (nonmetal)

4 0

3 years ago

X rays are used in hospital to locate in the patients bones.If the x rays of wavelength of 2×10 to the power negative nine m tra

jasenka [17]

Answer:1.5×10 to the power of 17(unit-Hertz/H)

Explanation:V=F×Wavelength

F=V/Wavelength=3×10 to power/2×10 to power of -9=1.5×10 to power of 17

8 0

3 years ago

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e

Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity $u=0$

acceleration $a=12.8 m/s^2$

velocity acquired by sled in $t_1$ time

$v=0+at$

$v=12.8t_1$

distance traveled by sled in $t_1 s$

$v^2-u^2=2as$

$(12.8t_1)^2-0=2\times 12.8\times s_1$

$s_1=6.4\cdot t_1^2$

distance traveled in $t_2$ time with velocity $v=12.8t_1$

$s_2=v\times t_2$

$s_2=12.8\times t_1\times t_2$

$s_2=12.8\cdot t_1\cdot t_2$

$s_1+s_2=5.37\times 10^3$

$6.4t_1^2+12.8t_1t_2=5370$ ----1

$t_1+t_2=97.7 s$

$t_2=97.7-t_1$

substitute the value of $t_2$ in 1

we get

$6.4t_1^2-1250.56t_1+5370=0$

thus $t_1=\frac{1250.56-1194.33}{12.8}=4.39 s$

$t_1=4.39 s$

5 0

3 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i

sammy [17]

Answer:

$-0.25 rad/s^2$

Explanation:

The equivalent of Newton's second law for rotational motions is:

$\tau = I \alpha$

where

$\tau$ is the net torque applied to the object

I is the moment of inertia

$\alpha$ is the angular acceleration

In this problem we have:

$\tau = -12.5 Nm$ (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

$I=50.0 kg m^2$ is the moment of inertia

Solving for $\alpha$ , we find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2$

3 0

3 years ago