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2 years ago

15

Thermal conductivity is a _______ property describing the material’s ability to conduct heat and is not dependent on the amoun

t of material present.

1 answer:

Elena-2011 [213]2 years ago

5 0

<h2>Answer: physical</h2>

Explanation:

Thermal conductivity is a physical property of materials that enable heat transmission. This means that when a body has thermal conductivity, heat passes from the body of higher temperature to another of lower temperature that is in direct contact with it.

In other words:

This property implies the ability of a material or substance to transfer the kinetic energy of its molecules to adjacent ones or to other materials (or substances) with which it is in contact.

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Fill in the blanks to correctly complete the statement. The motion of an object moving with uniform circular motion is always to

Citrus2011 [14]

The sentence can be completed as follows:

<span>The motion of an object moving with uniform circular motion is always tangential to the circle, so the speed of an object moving in a circle is known as tangential speed.

The object moves by uniform circular motion due to the presence of a force (called centripetal force) pointing toward the center of the circle. Due to the presence of this force, the object experiences an acceleration (called centripetal acceleration) that makes the object turning in a circle. This centripetal acceleration changes only the direction of the velocity of the object, not its magnitude, which is called tangential speed and it is constant.</span>

8 0

2 years ago

Read 2 more answers

Suzette had prepared the graph below to add to her lab

Charra [1.4K]

Answer:

A title

Explanation:

Because this is middle school.

4 0

2 years ago

Read 2 more answers

The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha

saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm = 4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0

2 years ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

4 0

2 years ago