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Kazeer [188]
3 years ago
12

Hrissie drinks cola every day. She texts the codes printed on the bottle caps to a reward center. For every 40 codes she texts,

she receives a free bottle of soda. Chrissie's behavior is reinforced on a _____ schedule.
Physics
2 answers:
andre [41]3 years ago
7 0
<span>Chrissie's behavior is reinforced on a FIXED-RATIO schedule.</span>
topjm [15]3 years ago
6 0

Answer: Fixed ratio schedule

Explanation:

 According to the question, chrissie behavior support the fixed ratio schedule as she text the message on every 40 sec and receive the free soda bottle. In the fixed ratio schedule is the type of schedule of the reinforced in which the response is given after some specific number of the response.

The fixed ratio schedule is a specific number and the fixed amount of the scheduled time. The fixed ratio is the constant that should be deliver with the constant number of the response that deliver correctly.  

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An object at rest starts accelerating.
Shalnov [3]

Answer:

<u>We are given: </u>

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

<u>Solving for 'a'</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0
2 years ago
A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
Elden [556K]

Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0
3 years ago
A 2.0-kilogram cantaloupe rolling north at 4.0 meters per second collides head on with a 1.0-kilogram orange rolling south at 8.
Murrr4er [49]

Answer:

Hi how are you doing today Jasmine

8 0
2 years ago
Read 2 more answers
Suppose you were asked to demonstrate electromagnetic induction. Which of the following situations will result in an electric cu
steposvetlana [31]
D. All of the above. When a wire loop is moved or rotated in a magnetic field, there is a change in magnetic flux which produces emf in wire loop and hence electric current is produced. 
5 0
3 years ago
Read 2 more answers
if 100 kilojoules of energy is used to heat 500 grams of water what is the temperature change of the water​
Serjik [45]

Answer:

47.8 °C

Explanation:

Use the heat equation:

q = mCΔT

where q is the heat absorbed/lost,

m is the mass of water,

C is the specific heat capacity,

and ΔT is the change in temperature.

Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.

100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT

ΔT = 47.8 °C

6 0
3 years ago
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