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m_a_m_a [10]
4 years ago
5

A rat receives food when it lever presses in the presence of a 1100 Hz tone, but not in the presence of a 1300 Hz tone. The 1100

Hz tone is an ________, while the 1300 Hz tone is an ________.
Physics
1 answer:
Paladinen [302]4 years ago
3 0

Answer:

1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )

Explanation:

This is used to describe Discrimination Training  in learning and behavior. The 1100 Hz is a discriminitive stimulus signal (SD), while 1300 Hz is a discriminitive stimuli for extinction (S∆ )

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When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequenc
Arlecino [84]

Answer:

m_2/m_1=9.745

The ratio m_2/m_1 of the masses is 9.745

Explanation:

Formula for frequency when mass m_1 is hung on the spring:

f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}

where:

k is the spring constant

Formula for frequency when mass m_2 is hung on the spring along with m_1:

f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}

where:

k is the spring constant.

In order to find ratio m_2/m_1, Divide the above equations:

\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}

On Solving the above equation:

\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745

The ratio m_2/m_1 of the masses is 9.745

8 0
4 years ago
If you punch a wall with a force of 30 N, what force will you feel on you fist? F= N
Talja [164]

Answer:

Pain

Explanation:

6 0
3 years ago
Read 2 more answers
The cheetah is one of the fastest accelerating animals, for it can go from rest to 28.0 m/s in 5.20 s. If its mass is 100 kg, de
Setler79 [48]

Answer:

a)15077 W

b)20.2185 horse power

Explanation:

P=F*V

F=ma

a=Vf-VS/t

Vf=28m/s

t=5.2

a=28/5.2

a=5.384 m/s²

F=100kg*5.384m/s²

F=538.4 N

P=F*V

P=538.4N*28m/s

P=15077 W=20.2185 horse power

1W=0.00134 Horse power

6 0
4 years ago
A steel rod with a length of l = 1.55 m and a cross section of A = 4.89 cm2 is held fixed at the end points of the rod. What is
Vedmedyk [2.9K]

Answer:

The size of the force developing inside the steel rod is 32039.28 N

Explanation:

Given;

length of the steel rod, L =  1.55 m

cross sectional area of the steel, A = 4.89 cm²

temperature change, ∆T = 28.0 K

coefficient of linear expansion for steel, α = 1.17 × 10⁻⁵ 1/K

Young modulus of steel,  E = 200.0 GPa.

Extension of the steel is given as;

α ∆T L = FL / AE

α ∆T = F/AE

F = AEα ∆T

F = ( 4.89 x 10⁻⁴)(200 x 10⁹)(1.17 × 10⁻⁵)(28.0 K)

F = 32039.28 N

Therefore, the size of the force developing inside the steel rod when its temperature is raised, is 32039.28 N

7 0
3 years ago
A mercury thermometer is constructed as
Tamiku [17]

Answer:

The change in height of the mercury is approximately  2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C

from which we solved for the change in volume \Delta V due to a given change in temperature \Delta T

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3

Therefore, the change in volume with a change in temperature of 36°C becomes:

\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

V_{cyl}=\pi r^2\,h\\h =\frac{V_{cyl}}{\pi r^2} \\h=\frac{0.0000474248\, cm^3}{\pi \, (0.00225\,cm)^2} \\h=2.98188 \,cm

8 0
3 years ago
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