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3 years ago

What is NOT a 21st-century skill that can help a person thrive today?

1 answer:

3 0

Ability to memorize and regurgitate information on tests is the skill that can’t help a person to thrive today in this modern 21st century.

Answer: Option D

<u>Explanation:</u>

Thriving in this modern 21st century is not as easy as it used to be in the early 20th and 19th centuries. So there is a need to develop skills like implementing one's knowledge to real-world structures which will sure help a person to understand and explore more.

Statement B also will sure be helpful to survive in modern society and a person can also improve his/her critical thinking, problem solving and other skills to thrive through this century.

But even though memorizing and regurgitating information can help you get good grade it will never be that helpful in other ways, So option D can be concluded as the right answer.

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Heating effect of current is not always useful for us. Support your answer with example<br>

stellarik [79]

Answer:

is correct

Explanation:

in my think, first this due to ray emitted from the light those ray may be affect our skin or party of body.

4 0

3 years ago

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position

antiseptic1488 [7]

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_{m}}$ ) is

$v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_{m}$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm$

6 0

3 years ago

A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical

Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water * ( T final - T equil ) = m ice* [c ice *( T equil -T initial ) + L + c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] + 1

since [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water > t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0

4 years ago

Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other in witch states coul

Stels [109]

Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?

Liquid is the answer

5 0

3 years ago

Read 2 more answers

Describe at least four of the requirements (facts) that any model of solar system formation must include (according to the nebul

kvasek [131]

The nebular theory describes the formation of the solar system and states that the system began as a gigantic cloud of gas and dust called a nebula which eventually condensed to form the sun, planets and other objects in the solar system. The first fact speaks to the formation of the planets, where gravity pulled larger clumps of material closer to form solid rocky planets closer to the sun and gas giants further out. The second requirement is that a nearby explosion or super nova would have to disturb our nebula to trigger rotation and the eventual formation of the sun. The third requirement/fact is that the planets go around the sun in the same direction. the last fact is that the planets go around the sun within 6 degrees of a common plane. This indicates that the solar system formed from a spinning disk of materials.

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3 years ago