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Lera25 [3.4K]
3 years ago
15

Juno created a diagram to describe characteristics of freezing and deposition. Which labels belong in the areas marked X and Y?

Physics
2 answers:
Fudgin [204]3 years ago
8 0

Answer:the correct answer is C

Explanation:

DIA [1.3K]3 years ago
7 0

the correct answer to this question is A

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Hull (1943) had rats push a lever that required 21 grams of force to budge. After they had learned to push the lever in order to
Zina [86]

Years of research have demonstrated that rats are intelligent creatures who experience pain and pleasure, care about one another, are able to read the emotions of others, and would assist other rats, even at their own expense.

<h3>Experiments:</h3>

In trials carried out at Brown University in the 1950s, rats were trained to press a lever for food, but they stopped pressing the lever when they noticed that with each press, a rat in an adjacent cage would scream in pain (after experiencing an electric shock).

Rats were trained to press a lever to lower a block that was hanging from a hoist by electric shocks administered by experimenters. A rat was subsequently hoisted into a harness by the experimenters, and according to their notes, "This animal normally shrieked and wriggled sufficiently while dangling, and if it did not, it was jabbed with a sharp pencil until it exhibited indications of discomfort." Even if it wasn't in danger of receiving a shock, a rat watching the scenario from the floor would pull a lever to lower the hapless rodent to safety.

Learn more about experiments on rats here:

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6 0
2 years ago
A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7
Norma-Jean [14]

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6


Therefore, v= 3.5 m/s.

7 0
3 years ago
Read 2 more answers
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
Two samples of dirt are collected from a suspect's tread in his shoe and a crime scene. She notes very similar characteristics.
elena-14-01-66 [18.8K]

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8 0
4 years ago
The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.
inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

         E = \frac{\sigma}{2 \times \epsilon_{o}}

            = \frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}

           = 1.18 \times 10^{5} N/C

Now, we will consider that equilibrium of forces are present there. So,

                   ma = qE

       a = \frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}

          = 1.13 \times 10^{13} m/s^2

According to the third equation of motion,

         v^{2} = 2 \times a \times d

or,      d = \frac{v^{2}}{2d}

             = \frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}

             = 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0
3 years ago
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