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Savatey [412]
3 years ago
8

A mother pats a child every time he throws a chocolate wrapper in the dustbin. This is an example of (A) Observational Learning

(B) Negative Reinforcement (C) Spontaneous recovery (D) Positive Reinforcement
Physics
1 answer:
TEA [102]3 years ago
8 0
D. I think, not sure
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Which is an example of a physical change?
Ber [7]

Answer:

salt dissolving in water

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A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?
maria [59]

Answer:

12500W

Explanation:

Given parameters:

Work done  = 250000J

Time taken  = 20s

Unknown:

Power of the crane = ?

Solution:

Power is the defined as the rate at which work is being done;

  Mathematically;

        Power = \frac{work done}{time }

 insert the parameters and solve;

      Power  = \frac{250000}{20}   = 12500W

7 0
3 years ago
Which of the following statements is true?
hram777 [196]

All the radiation from stars is the result of nuclear fusion in their cores.

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3 years ago
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In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
3 years ago
What is the law of variation of the period T of a simple pendulum
DIA [1.3K]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum and g=9.81 m/s^2 is the gravitational acceleration. As we can see, the period of a simple pendulum depends only on its length.
3 0
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