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3 years ago

Which of the following is true of education in 1950

1 answer:

8 0

Information I learned from history class Education in the 1950's expanded from previous decades. They no longer focused purely on reading, writing and arithmetic. History and science became a main part of the cirriculum. Also, enrollment skyrocketed as the baby-boomers began enrolling in elementary school. One interesting thing that categorized this generation was the presence of fallout tests. Schools would require the students to go through a fake atomic bomb attack in which they would hide under their desks (which was completely pointless in protecting them from radiation, it was more of an emotional security for the parents and teachers, but scared the hell out of the students). Socially, children were taught to conform and to be normal. Standing out or questioning authority was bad. Sex was taught, though minimally. They explained the penis and vagina. Sexually transmitted diseases were focused on greatly so as to "scare" the students out of premarital sex.

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What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m

masya89 [10]

Answer: $0.0058 m/s^{2}$

Explanation:

Centripetal acceleration $a_{C}$ is calculated by the following equation:

$a_{C}=\frac{V^{2}}{r}$

Where:

$V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s$ is the Earth's orbital speed

$r=1.5(10)^{11} m$ is the orbital radius

$a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}$

$a_{C}=0.0058 m/s^{2}$

4 0

3 years ago

An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0

3 years ago

In what ways does a stream transport its load, and which part of the load moves most slowly? Select all that apply.

vodka [1.7K]

Answer: Option (B)

Explanation: A stream transports its materials in different ways-

Dissolved load- Here, the materials gets dissolved when mixed with water and flows along with the stream.
Suspended load- Here, the materials are not fully dissolved in the water but they can be carried from one place to another in suspension mode, by the river.
Bed load- Bed load are transported in three different ways such as-

Sliding- here, the materials slides down along a curved surface under the water and carried away.
Rolling- here, the materials are solid and due to force exerted by water, it can roll and move to distant places.

Saltation- here, the materials are carried away in a series of jumps.

Thus, the most appropriate answer is option (B) i.e bedload.

5 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$