What is silviculture?

PLEASE HELP!!!

PtichkaEL [24]

Answer:

Densities increase down the group

MP and BP decrease down the group

Softness increased going down the group

Speed of reacting increases going down the group

4 0

3 years ago

Metals are used in many products because of the characteristic properties that most metals share. Which properties of the metal

Anastaziya [24]

The most important characteristics that are exhibited by metals are-

1- Metals are ductile

2-Most metals are conductive in nature.

3-Most metals are malleable.

4- Metals have strong inter molecular force of attraction between the.

5-Metals have luster.

6-Metals are sonorous.

Here we are given Tungsten filament.

Tungsten is a metal.So it must be conductive and as well as ductile in nature.

The electric filament that we are using in our electric bulb glows due to the heating effect of current.Hence the chosen substances for glowing electric bulb must have high melting point.

The melting point of tungsten is 1650 degree celsius which is very high.That's why it is used in electric bulb.

Hence the correct answer to the question is the third one i.e Tungsten is ductile,has a high melting point, and is electrically conductive.

5 0

3 years ago

Read 2 more answers

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

7 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago

6. A child tugs on a rope attached to a 0.62-kg toy with a horizontal force

ser-zykov [4K]

Answer:

Explanation:

Force Mass * acceleration

F = ma

a = F/m

Sum of force = 16.3N - 15.8N = 0.5N

Mass = 0.62kg

Substitute

a = 0.5/ * 0.62

a = 0.81m/s²

The acceleration of the toy is 0.81m/s²

8 0

3 years ago