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3 years ago

How do the earth's physical features affect people's activities?

Physics

1 answer:

masya89 [10]3 years ago

4 0

Answer:

d.all of the above

Explanation:

The physical feature of the planet Earth affect people's activities in the following manner -

Choice of Occupation - the process of occupation is completely a land depended process , hence any changes on the physical features of Earth , like the climate , temperature , rain etc , will hinder the occupations process .
Method of transportation - it is also a Climate dependent process , so any change in climate may obstruct the transportation facility .

Recreational activities - also depends on the physical feature of the planet Earth .

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Exercise provides a healthy outlet for feelings, which helps improve __________.

stiks02 [169]

A. Emotional Health
Hope this helps

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3 years ago

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A chain reaction results when a uranium atom is struck by a/an ______________released by a nearby Uranium atom undergoing fissio

finlep [7]

Answer:

Chain reaction is possible by neutron

Explanation:

Nuclear reaction is mainly two types,

⇒ Nuclear Fission : heavy atom split into two light atom.

Ex. Uranium, thorium

⇒ Nuclear fusion : lighter atom combine together

Ex. Hydrogen to helium

In fusion reaction the large amount of energy is produced as compare to fission reaction.

Sun gets brighter by fusion reaction.

In case of uranium fission reaction is possible by colliding neutron.

8 0

4 years ago

(1 point) Suppose a spring with spring constant 1 N/m is horizontal and has one end attached to a wall and the other end attache

fredd [130]

Answer:

Explanation:

Given that,

The spring constant

K = 1N/m

Frequency of motion

f = 14Hz

We want calculate the mass m?

The frequency of spring system is related to the mass by

From w = √k/m

Where w = 2πf

f = 1/2π √k/m

Where,

w is angular frequency in rad/s

m is mass of object attached in kg

k is the spring constant in N/m

f Is the frequency in Hz

Then, make m subject of formula

Multiply both sides by 2π

2πf = √k/m

Square both sides

4π²f² = k/m

Then, k= 4π²f² × m

m = k / 4π²f²

m = 1 / (4π² × 1.4)

m = 1 / 55.27

m = 0.0181 kg

m= 18.1 g

The mass of the object attached in 18.1 g or 0.0181 kg

4 0

3 years ago

Definition of wave velocity

maksim [4K]

Answer:

Velocity = frequency * wave length

Explanation:

Wave velocity is the measure of how fast the wave is being transmitted to a particular direction.

velocity implies both speed and direction as we know.

The velocity of a wave is equal to the product of its wavelength and frequency

where

frequency - number of oscillations per unit time (measured in Hz)
wave length - distance between corresponding points of two consecutive waves (measured in meters)

3 0

3 years ago

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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 2.33 m away from a waterfall 0.488 m in heigh

Ulleksa [173]

Answer:

5.68 m/s

Explanation:

The motion of the salmon is the same as a projectile: it is launched with an initial speed $u$ at an angle of $\theta=38^{\circ}$ above the horizontal.

The motion of the salmon consists of two indipendent motion:

- Along the horizontal direction, it is a uniform motion with constant velocity

$v_x = u cos \theta$

So that the distance travelled is

$d=v_x t = u cos \theta t$ (1)

- Along the vertical direction, it is a uniformly accelerated motion with constant acceleration downward, so the vertical displacement is

$y = u sin \theta t - \frac{1}{2}gt^2$ (2)

where g is the acceleration of gravity.

We know the following:

- The horizontal distance travelled by the salmon to reach the waterfall is

d = 2.33 m

- The vertical distance travelled is the height of the waterfall,

y = 0.488 m

From (1) we get:

$t=\frac{d}{u cos \theta}$

And substituting into (2), we can solve the equation to find t, the time at which the salmon reaches the waterfall:

$y = u sin \theta \frac{d}{u cos \theta} - \frac{1}{2}gt^2 = d tan \theta - \frac{1}{2}gt^2\\t = \sqrt{\frac{2(d tan \theta - y)}{g}}=\sqrt{\frac{2(2.33 tan 38^{\circ}-0.488)}{9.81}}=0.521 s$

And then, we can use eq.(1) again to find the initial speed, u:

$u=\frac{d}{cos \theta t}=\frac{2.33}{cos(38^{\circ})(0.521)}=5.68 m/s$

5 0

3 years ago