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3 years ago

6

What role does a mild electric shock play in Skinner's operant chamber? A. negative reinforcer B. positive reinforcer C. punishe

r D. neutral operant

1 answer:

photoshop1234 [79]3 years ago

8 0

Answer: b

Explanation:

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If a force of 24 N is applied to an object with a mass of 5 kg, the object will accelerate at _____ m/s2

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4.8m/s2 according to the formula F=MA

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3 years ago

A cannon ball is fired with an initial velocity of 75 m/s at an angle of 58 degree above the ground. What maximum height will it

KiRa [710]

Answer:

h = 206.4 m

range = 515.9 m

Explanation:

from the question we are given the following:

initial velocity (u) = 75 m/s

angle above surface = 58 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

find the maximum height (h) and the horizontal distance

maximum height (h) = $\frac{u^{2}sinθ^{2} }{2g}$

h = $\frac{75^{2}sin58^{2} }{2 x 9.8}$

h = 206.4 m

the horizontal distance here is the range

range = $\frac{u^{2}sin2θ }{g}$

range = $\frac{75^{2}sin(2 x 58) }{9.8}$

range = 515.9 m

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3 years ago

Which describes an image that can be produced by a concave lens?

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4 years ago

Explain how there can be more heat energy in an iceberg than in a pot of boiling water

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4 years ago

Read 2 more answers

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

4 years ago