Describe what “melting point” means

Which of the following is the best thermal insulator?

serious [3.7K]

Answer:

A. Air

Explanation:

A thermal insulator is a material that does not let the heat trasmitting through it, or it transmits heat not very efficiently.

Generally, the lower the density of the material, the better insulator the material is: this is because in materials with very low density, particles are very spread apart, so they cannot collide with each other, and so the vibrations of the particles (which determine how hot a substance is) cannot be transmitted very efficiently.

In the options listed in this problem, air is the material with lowest density, therefore it is the best thermal insulator among them.

3 0

3 years ago

Read 2 more answers

Name the physical quantity measured by the velocity-time graph?

VARVARA [1.3K]

The so-called "velocity-time" graph is actually a "speed-time" graph. At any point
on it, the 'x'-coordinate is a time, and the 'y'-coordinate is the speed at that time.

'Velocity' is a speed AND a direction. Without a direction, you do not have a velocity,
and these graphs never show the direction of the motion. It seems to me that it would be
pretty tough to draw a graph that shows the direction of motion at every instant of time,
so my take is that you'll never see a true "velocity-time" graph.

At best, it would need a second line on it, whose 'y'-coordinate referred to a second
axis, calibrated in angle and representing the 'bearing' or 'heading' of the motion at
each instant. The graph of uniform circular motion, for example, would have a straight
horizontal line for speed, and a 'sawtooth' wave for direction.

7 0

3 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

2 years ago

Explain the relationship between resistance and the velocity of shortening.

butalik [34]

The velocity of shortening refers to the speed of the contraction from the muscle shortening while lifting a load. The relationship between the resistance and velocity of shortening is inverse. The greater the resistance, the shorter the velocity of shortening and the smaller the resistance, the larger the velocity of shortening.

Hopefully this help :)

7 0

3 years ago

A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.

Pavel [41]

Hopefully this will help you.

4 0

3 years ago