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3 years ago

11

When he sees teachers encouraging other children to wait in the cafeteria until the first bell rings, Ian follows them. What typ

e of learning is Ian demonstrating?

2 answers:

Ostrovityanka [42]3 years ago

8 0

This is observational learning because Ian observed that his peers waited in the cafeteria until the first bell rings. He decided to imitate them.

Tresset [83]3 years ago

8 0

ANSWER: Ian is demonstrating a typical example of Observational Learning. An observational learning is a learning pattern which happens by observing the behavior of others. People generally follows a role model such as parent, sibling, relative or teachers and imitate them while they are doing their work, particularly in childhood. There are four steps of observational learning which are Attention, Memory, Initiation and Motivation.

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An astronaut in her space suit has a total mass of 87.0kg including suit and oxygen tank. Her tether line loses its attachment t

Shtirlitz [24]

Answer:

Explanation:

a )

In space due to weightlessness both astronaut and her oxygen tank will float .

when she throws the tank away from spacecraft , she will have a velocity in opposite direction ie towards the spacecraft . This happens due to conservation of momentum . She creates a momentum away so that she can get a momentum towards the spaceship.

So

m₁ v₁ = m₂v₂

12 x 8 = ( 87 - 12 ) x v₂

v₂ = 1.28 m /s

Time allowed = 2 x 60

= 120 s

So maximum distance upto which she can remain away from spacecraft

= 120 x 1.28

= 153 m .

b )

The Newton's law which explains the theory behind it is "third law of motion" . This law gives law of conservation of momentum .

6 0

4 years ago

A 200 g, 20 cm diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supp

Rama09 [41]

Answer: The torque required is 0.0471 N m.

Explanation:

Mass of the disc = 200 g = 0.2 kg (1 kg =1000 g)

Radius of the disc = $\frac{Diameter}{2}$ = 10 cm = 0.1 m(1 m = 100 cm)

Angular acceleration = $\alpha$

$\alpha =\frac{2\pi\times 1800}{60 \times 4 sec}=15\pi rad/s^2$

Moment of inertia = $\frac{1}{2}\times mass\times (radius)^2=\frac{1}{2}\times0.2 kg\times (0.1 m)^2=0.001 kg m^2$

$Torque=\alpha \times I=0.001 kg m^2\times 15\times 3.14 rad/s^2r=0.0471 N m$

The torque required is 0.0471 N m.

7 0

4 years ago

In uniform circular motion, which of the following are constant:Check all that apply.Check all that apply.the speed is constantt

AlladinOne [14]

Answer:

The speed, magnitude of the velocity, magnitude of the angular velocity, magnitude of the centripetal acceleration, magnitude of the net force and direction of the angular velocity are constant.

Explanation:

In uniform circular motion we have a centripetal acceleration of constant magnitude but changing direction (since it points to the center of the circle from the object). The same goes for the net (centripetal) force since F=ma. This makes the magnitude of the velocity (speed) constant but its direction changes, although keeping spinning in the same direction, which makes its angular velocity constant in both magnitude and direction.

3 0

4 years ago

Read 2 more answers

Earth-orbiting astronauts feel weightless in space because _____. Choose all that apply. 1 point They are in free-fall motion. T

STALIN [3.7K]

Answer:

They are in free-fall motion.

Explanation:

The Earth orbiting astronauts are falling at an acceleration that is the same or greater than the acceleration due to gravity i.e., 9.81 m/s². If you are continuously falling at this rate then you will feel weightless.

This same effect is felt while going down in an elevator. When you down in an elevator you feel that you are lighter and feel that something is pushing you up. Earth-orbiting astronauts feel the same effect but the accelration is greater hence they feel weightless.

5 0

3 years ago

A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity

nadezda [96]

Answer:

$u(t)=\frac{1}{5} sin\ (25t)$

Explanation:

Given:

mass of the body stretching the spring, $m=150\ g$

extension in spring, $\Delta x=1.568\ cm$

velocity of oscillation, $u'(0)=20\ cm.s^{-1}$

initial displacement position of equilibrium, $u(0)=0$

<u>According to given:</u>

$m.g=k.\Delta x$

$150\times 980=k\times 1.568$

$k=93750\ dyne.cm^{-1}$

<u>we know frequency:</u>

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{93750}{150} }$

$\omega=25$

Now, for position of mass in oscillation:

$u= A.sin\ (\omega.t)+B.cos\ (\omega.t)$

$u= A.sin\ (25.t)+B.cos\ (25.t)$

at $t=0;\ u(0)=0\ \Rightarrow A=0$

∴ $u(t)=B.sin\ (25.t)$

∵ at $t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}$

$u(t)=\frac{1}{5} sin\ (25t)$

7 0

3 years ago