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2 years ago

What disagreement did Sigmund Freud have with both Josef Breuer and Jean Martin Charcot?

Physics

1 answer:

GREYUIT [131]2 years ago

4 0

Answer:

The correct answer is 'A'

Explanation:

I guessed and was correct.

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Can someone please help a struggling physics student?

zhenek [66]

<u><em>What is the maximum height the ball will reach in the air?</em></u>

Kinematics equation used:

$v_f^2=v_i^2+2ad$ , where $v_f$ is final velocity, $v_i$ is initial velocity, $a$ is acceleration, and $d$ is distance travelled. From SI units, velocity should be in $m/s$ , acceleration should be in $m/s^2$ , and distance should be in $m$

We're given that the initial velocity is 12.0 m/s in the y-direction. At the maximum height, the vertical velocity of the ball will be 0 m/s, otherwise it would not be at maximum height. This is our final velocity.

The only acceleration in the system is acceleration due to gravity, which is approximately $9.8\:\mathrm{ m/s^2}$ . However, the acceleration is acting down, whereas the ball is moving up. To express its direction, acceleration should be plugged in as $-9.8\:\mathrm{m/s^2}$ . We have three variables, and we are solving for the fourth, which is distance travelled. This will be the maximum height of the ball.

Substitute $v_i=12$ , $v_f=0$ , $a=-9.8$ to solve for $d$ :

$0^2=12^2+2(-9.8)(d),\\0=144-19.6d,\\-19.6d=-144,\\d=\frac{-144}{-19.6}=7.34693877551\approx \boxed{7.35\text{ m}}$

<u><em>What is the velocity of the ball when it hits the ground?</em></u>

This question tests a physics concept rather than a physics formula. The vertical velocity of the ball when it hits the ground is equal in magnitude but opposite in direction to the ball's initial vertical velocity. This is because the ball spends equal time travelling to its max height as it does travelling from max height to the ground (ball is accelerating from initial velocity to 0 and then from 0 to some velocity over the same distance and time). Since the ball has an initial vertical velocity of +12.0 m/s, its velocity when it hits the ground will be $\boxed{-12.0\text{ m/s}}$ . (The negative sign represents the direction. Because velocity is a vector, it is required.)

<u>**Since my initial answer exceeds the character limit, I've attached the first question to Part B as an image. Please refer to the attached image for the answer and explanation to the first question of Part B. Apologies for the inconvenience.**</u>

<u><em>What is the direction of the velocity of the ball when it hits the ground? Express your answer in terms of the angle (in degrees ) of the ball's velocity with respect to the horizontal direction (see figure).</em></u>

This question uses a similar concept as the second question of Part A. The vertical velocity of the ball at launch is equal in magnitude but opposite in direction to the ball's final velocity. The horizontal component is equal in both magnitude and direction throughout the entire launch, since there are no horizontal forces acting on the system. Therefore, the angle below the horizontal of the ball's velocity when it hits the ground is equal to the angle of the ball to the horizontal at launch.

To find this, we need to use basic trigonometry for a right triangle. In any right triangle, the tangent/tan of an angle is equal to its opposite side divided by its adjacent side.

Let the angle to the horizontal at launch be $\theta$ . The angle's opposite side is represented by the vertical velocity at launch (12.0 m/s) and the angle's adjacent side is represented by the horizontal velocity at launch (2.3 m/s). Therefore, we have the following equation:

$\tan \theta=\frac{12.0\text{m/s}}{2.3\text{ m/s}}$

Take the inverse tangent of both sides:

$\arctan (\tan \theta)=\arctan (\frac{12.0}{2.3})$

Simplify using $\arctan(\tan \theta)=\theta \text{ for }\theta \in (-90^{\circ}, 90^{\circ})$ :

$\theta=\arctan(\frac{12.3}{2.3}),\\\theta =79.14989537\approx \boxed{79.15^{\circ}}$

We can express our answer by saying that the direction of the velocity of the ball when it hits the ground is $\boxed{\text{approximately }79.15^{\circ} \text{ below the horizontal}}$ or $\boxed{\text{approximately }-79.15^{\circ} \text{ to the horizontal}}$ .

5 0

3 years ago

Which segment of the graph represents constant speed? Explain your answer.

taurus [48]

D is the answer

in V-t graph, tan α= velocity/time=acceleration

7 0

3 years ago

Some believe that the positions of the planets at the time

laila [671]

Answer:

$1.4007\times 10^{-8}\ N$

$1.50075\times 10^{-6}\ N$

$0.000000667\ N$

Explanation:

$m_1$ = Mass of baby = 3 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Distance between objects

Gravitational force of attraction is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 70\times 3}{1^2}\\\Rightarrow F=1.4007\times 10^{-8}\ N$

The force between baby and obstetrician is $1.4007\times 10^{-8}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(6\times 10^{11})^2}\\\Rightarrow F=1.50075\times 10^{-6}\ N$

The force between the baby and Jupiter is $1.50075\times 10^{-6}\ N$

$F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(9\times 10^{11})^2}\\\Rightarrow F=0.000000667\ N$

The force between the baby and Jupiter is $0.000000667\ N$

7 0

3 years ago

A trough is 10 feet long and its ends have the shape of isosceles triangles that are 3 feet across at the top and 1 foot high. I

tensa zangetsu [6.8K]

Answer:

1.6 ft/min

Explanation:

Since trough is 10 ft long and water is filled at the rate of 12ft3/min. We can calculate the rate of water filled with respect to area:

= 12 / 10 = 1.2ft2/min

As the water level rises, so does the water surface, or the bottom side of the isosceles triangles. In fact we can calculate the bottom side when the trough is half foot deep:

= 3 / 2 = 1.5 ft

The rate of change in water level would be the same as calculating the height of the isosceles triangles knowing its base

= 1.2 * 2 / 1.5 = 1.6 ft/min

6 0

3 years ago

Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has

Degger [83]

Answer:

Four charges of equal magnitude sitting at the vertices of a square

Explanation:

We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.

Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.

We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.

6 0

2 years ago