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pav-90 [236]
3 years ago
11

Volcano has both useful and harmful effects give reason​

Physics
2 answers:
Mama L [17]3 years ago
6 0

Answer:

useful effects of volcano are :-

  1. it makes soil fertile
  2. it provides valuable nutrients for the soil

harmful effects of volcano are:-

  1. it makes air polluted
  2. it destroy the environment .

hope it is helpful to you ☺️

alina1380 [7]3 years ago
3 0

Answer:

harmful effects

1. that will cause air pollution

2. that will destroy our earth

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A hockey puck is hit on a frozen lake and starts moving with a speed of 13.60 m/s. Exactly 6.2 s later, its speed is 7.20 m/s. (
stellarik [79]

Answer:

-1.03 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².

Mathematically, acceleration is expressed as

a = (v-u)/t ........................ Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t  = time.

Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.

Substituting into equation 2

a = (7.20-13.60)/6.2

a = -6.4/6.2

a = -1.03 m/s²

Note: a is negative because, the hockey puck is decelerating.

Hence the average acceleration = -1.03 m/s²

3 0
2 years ago
Two concrete spans of a 380 m long bridge are
Mazyrski [523]

Answer:

4.163 m

Explanation:

Since the length of the bridge is

L = 380 m

And the bridge consists of 2 spans, the initial length of each span is

L_i = \frac{L}{2}=\frac{380}{2}=190 m

Due to the increase in temperature, the length of each span increases according to:

L_f = L_i(1+ \alpha \Delta T)

where

L_i = 190 m is the initial length of one span

\alpha =1.2\cdot 10^{-5} ^{\circ}C^{-1} is the temperature coefficient of thermal expansion

\Delta T=20^{\circ}C is the increase in temperature

Substituting,

L_f=(190)(1+(1.2\cdot 10^{-5})(20))=190.0456 m

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

h=\sqrt{L_f^2-L_i^2}=\sqrt{(190.0456)^2-(190)^2}=4.163 m

4 0
3 years ago
How do scientists know our universe is older than 6500 years?
dmitriy555 [2]
<span>This is because astronomers have discovered that the Crab Nebula is 6500 light years from earth. This means it takes 6500 earth years for light to get from the Crab Nebula to earth. Therefore, if we are able to observe the Crab Nebula and beyond it then it means that earth has been in existence for 6500 years and over.  </span>




5 0
2 years ago
19. State any 3 applications of capillary action (3mk)​
zlopas [31]

Explanation:

1. Movement of water, food and mineral salts in plants

2. Absorption of water by towels when wiping our bodies

3. It is used to absorb ink using a blotting paper or tissue

7 0
3 years ago
Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0
3 years ago
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