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3 years ago

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Which two stimuli did John B. Watson associate in his infamous “Little Albert” experiment? A. a white lab rat and the boy’s moth

er B. fear and a loud noise C. a white lab rat and a loud noise D. fear and the boy’s mother

1 answer:

MA_775_DIABLO [31]3 years ago

7 0

I got answer c but im not 100% sure

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The mass of a cannon ball is 10 kg. If the speed of the cannon ball is 50 m/s in

Maurinko [17]

Answer:

Explanation:

Kinetic Energy = 0.5(Mass)(Velocity2)

Kinetic energy= 0.5 × 10kg × (50m/s)2

Kinetic Energy = 5kg × 2500m/s

Kinetic energy = 125000 J ( Ans)

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3 years ago

!!!!!!PLEASE HELP ASAP !!!!!!!!!!!!

OLga [1]

Answer: d constint speed

Explanation: im 15 and no the answer

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3 years ago

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$

Using (1),

When $\rm q_1 = \pm 8.00\times 10^{-7}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.$

When $\rm q_1=\pm 4.6\times 10^{-6}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.$

Since, $\rm |q_2|>|q_1|$

Therefore, $\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

7 0

3 years ago

Which one of the following environments will sound not travel through?

Eduardwww [97]

The answer is d. Vacuum

Sound waves need a molecul to travel through. Gas , solid, annd liquid provide that medium.

But vacuum is just an empty space withour any moleculs in it.

Thats why in space, no one can hear you scream. Because it simply not physicsly possible

7 0

3 years ago

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

Marina CMI [18]

Answer:

(a) $\vec{J}=176.4\frac{kg*m}{s} \hat{j}$

(b) $F=2125.30N$

Explanation:

(a) According to the law of conservation of energy, the potential energy of the person at 0.40 m is equal to its kinetic energy before the colision with the floor:

$\Delta U=\Delta K\\mgh=\frac{mv^2}{2}\\v=\sqrt{2gh}\\v=\sqrt{2(9.8\frac{m}{s^2})(0.40m)}\\v=2.8\frac{m}{s}$

This is the initial velocity in the negative y-direction. Impulse is given by:

$\vec{J}=\Delta \vec{p}\\\vec{J}=m\vec{v_f}-m\vec{v_i}\\\vec{J}=63kg(0 \hat{j})-63kg(-2.8\frac{m}{s} \hat{j})\\\vec{J}=176.4\frac{kg*m}{s} \hat{j}$

(b) The average force is:

$F=\frac{J}{\Delta t}\\F=\frac{176.4\frac{kg*m}{s}}{0.083s}\\F=2125.30N$

6 0

3 years ago