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2 years ago

What did Wundt's pendulum experiment demonstrate?

Physics

2 answers:

yawa3891 [41]2 years ago

8 0

Answer:

C) People need time to shift between two different stimuli.

Explanation:

Wilhelm Wundt attempted to study the behavior of mind by measuring and observing the body. He conducted an experiment which would measure the rate of speed of thoughts of the subjects to make any judgment. He experimented by measuring the tie with a pendulum swing. He is referred to as the 'father of experimental psychology' because of his efforts in the field of psychology.

weeeeeb [17]2 years ago

6 0

C. <span>People need time to shift between two different stimuli.</span>

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Heat flows into a gas in a piston and work is performed on the gas by its surroundings. The amount of work done is equal to the

inna [77]

Answer:

The Internal energy of the gas did not change

Explanation:

In this situation the Internal energy of the gas did not change and this is because according the the first law of thermodynamics

Δ U = Q - W ------ ( 1 )

Δ U = change in internal energy

Q = heat added

W = work done

since Q = W. the value of ΔU will be = zero i.e. No change

4 0

2 years ago

a light-year is ____ A. the amount of time it takes light to get to the nearest star B. the distance light travels in a day C. t

Vinil7 [7]

The Answer is C. the distance light travels in a year

6 0

2 years ago

Read 2 more answers

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago

Two sound waves, A and B, are traveling at the same speed. Wave A has a wavelength of 50 cm and a frequency of 7000 Hz. Wave B h

vekshin1

Answer:

D. 100 cm

Explanation:

The speed of a wave is the wavelength times the frequency.

v = λf

Wave A and B have the same speed, so:

λf = λf

(50 cm) (7000 Hz) = λ (3500 Hz)

λ = 100 cm

6 0

2 years ago

THIS MARCIN

nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

−

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

′

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

′

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

−

(−25)

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

(−25)

−

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

250

25−10

250

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

250

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

′

=−

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0

2 years ago