Justina dreamed last night that she was warding off villains in a life or death battle outside a fortified castle. In the dream, she attempted to cross the moat but saw that it was filled with grotesque swamp creatures with warty green skin and she recoiled in fear. Knowing there was only one way to reach safety, she flew over the top of the castle and landed among the weeds in the inner courtyard. She was surprised to run into her seventh period Geometry teacher, Ms. Hargroves, but seeing that her teacher had a spear, compass, and workbook with her, Justina knew they would prevail. Ms. Hargroves told Justina that the key to defeating the villains was to write the theorems from the homework on the castle walls as protection against the invaders. Justina scrambled to write the complicated theorems before the menacing villains closed in. Just when the towering castle walls were about to be sieged, Justina awoke from her dream with a start.

As a therapist who specializes in dreams, who would you interpret Justina’s dream using:
17. Freud’s wish-fulfilling theory:
18. The information-processing theory:
19. The physiological function theory:
20. The activation synthesis theory:
21. The cognitive development theory:

The momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.

The momentum transfered to the more massive cart is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

⁻ΔP₁ = ΔP₂

ΔP₂ = m₂v₂ - m₂u₂

ΔP₂ = m₂(v₂ - u₂)

ΔP₂ = 3m₁(v₂ - u₂)

ΔP₂ = 3 x 3.8 x (1.7 - 0)

ΔP₂ = 19.38 kgm/s

Thus, the momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.

The complete question is beblow

A cart of mass 3.8 kg is traveling to the right (which we will take to be the positive x-direction for this problem) at a speed of 6.9 m/s. It collides with a stationary cart that is three times as massive. After the collision, the more massive cart is moving at a speed of 1.7 m/s, to the right.

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Answer:

M1 g L1 = 19 kg * 9.8 m/s^2 * 5 m = counter clockwise torque - Cassie at left end

M1 g L1 = M2 g L2        for torques to balance

L2 = M1 L1 / M2 = 19 * 5 / 31 = 3.06 M

Alex should sit at 3.1 m from the fulcrum (at 5 m from each end)

The monocular cue of relative size

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

Answer:

If one wraps the fingers around the wire and points the thumb in the direction of the "conventional" current the fingers will point towards the North pole - the direction of the B-field.

In this case the B-field is pointed "West".

Answer:

M1 would seem to be slower because of a larger mass

x1 = A1 sin ω1 t1        describes the displacement

ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2  since k's are equal

ω1 / ω2 = 1/2 from graph    (frequency of 2 is greater)

(m1 / m2)^1/2 = ω2 / ω1    from above

m1 / m2 = 2^2 = 4    so m1 would have 4 times the mass of m2

M1 would seem to be slower because of a larger mass

x1 = A1 sin ω1 t1      

ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2  since k's are equal

ω1 / ω2 = 1/2 from graph   (frequency of 2 is greater)

(m1 / m2)^1/2 = ω2 / ω1   from above

m1 / m2 = 2^2 = 4    so m1 would have 4 times the mass of m2.

The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system. For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Therefore, M1 would seem to be slower because of a larger mass

x1 = A1 sin ω1 t1      

ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2  since k's are equal

ω1 / ω2 = 1/2 from graph   (frequency of 2 is greater)

(m1 / m2)^1/2 = ω2 / ω1   from above

m1 / m2 = 2^2 = 4    so m1 would have 4 times the mass of m2.

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Answer:

a

Explanation:

Distance is simply the distance travelled which in this case would be 4km + 3km = 7km

To work out displacement, try to imagine the situation.

Draw a straight line to the east (label it 3) and then draw another line from the end of the first line upwards (label this one 4). Thus, you've created a right angles triangle. Now use pythagorean theorem to work out the displacement

4^2 + 3^2 = 25

sqrt 25 = 5 = displacement

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

Answer:

5.98 kg

Explanation:

To solve this problem, let use the Linear Momentum Conservation Law:

Before collision: [tex]\sum p=p_{1}+p_{2}=m_{1}v_{1}+m_{2}v_{2}=(2.99v_{1})+0=2.99v_{1}[/tex]

After collision: [tex]\sum p'=p_{1}'+p_{2}'=(2.99+m_{2})(v_{1}/3)[/tex]

So, we obtain:

[tex]\sum p =\sum p' \rightarrow 2.99v_{1}=(2.99+m_{2})(v_{1}/3) \rightarrow 8.97 = 2.99 + m_{2}[/tex]

[tex]m_{2}=8.97-2.99=5.98 kg[/tex]

If a person generates about 412005 joules of heat from the body,  the water temperature is mathematically given as

t=21.6296C

Question Parameter(s):

The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C

Generally, the equation for the Heat   is mathematically given as

Heat gained =Heat loess

Thereofore

mw*cw*(t-2160)=1.5*10^5

[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]

t=21.6296C

In conclusion, the tempreature

t=21.6296C

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Answer:

[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)

Explanation:

The forces on the bucket are:

Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:

[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].

The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].

Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:

[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].

At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].

Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].

Therefore, the magnitude of the angular acceleration of this sphere would be:

[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].

The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].

Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].

Hence the equation:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].

Solve this equation for [tex]M[/tex], the mass of this sphere:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].

[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].

[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].

Answer:

Explanation:

Speed of light v = 3 x 10⁸ m/s

wavelength λ = 8.1 x 10⁻⁸ m

frquency f = v/λ = 3.7 x 10¹⁵ Hz

If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

C = λν

As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,

The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸

                                            =3.7 × 10¹⁵ Hz

Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz

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Answer:

Given:

m1 = 7 kg

V1 = 12 m/s

m2 = 25 kg

V2 = 6 m/s

To find:

Combined speed of two balls stick together after collision V = ?

Solution:

According to law of conservation of momentum,

m1V1 + m2V2 = (m1+m2)V

7×12 + 25×6 = (7+25)V

84 + 150 = 32V

V = 234/32

V = 7.31 m/s

Combined speed of two ball is 7.31 m/s

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Answer:

Step by step explanation:

Step 1: Define an equation that relates the volume of a sphere to its radius.

V = 4/3*π*r3

Step 2: Take the derivative of each side with respect to time (we will define time as "t").

(d/dt)V = (d/dt)(4/3*π*r3)

dV/dt = 4πr2*dr/dt

Step 3: We are told in the problem statement that diameter is 100m, so therefore r = 50mm. We are also told the radius of the sphere is increasing at a rate of 2mm/s, so therefore dr/dt = 2mm/s. We are looking for how fast the volume of the sphere is increasing, or dV/dt.

dV/dt = 4π(50mm)2*(2mm/s)

dV/dt = 62,832 mm3/s

(a) The initial energy of the block due to its position is 1.96 J.

(b) The velocity of the block at the bottom of the loop is 3.96 m/s.

(c)  the velocity of the block at the top of the loop is 3.13 m/s.

The initial energy of the block due to its position is calculated as follows;

P.E = mgh

P.E = 0.25 X 9.8 X 0.8

P.E = 1.96 J

The velocity of the block at the bottom of the loop is determined by applying the principle of conservation of energy as shown below;

P.Ei + P.Ef = K.Ei + K.Ef

1.96 + 0 = 0 + ¹/₂mvf²

vf² = 2(1.96)/m

vf² = (2 x 1.96) / (0.25)

vf² = 15.68

vf = √15.68

vf = 3.96 m/s

The velocity of the block at the top is calculated by applying principle of conservation of energy,

P.Ei + P.Ef = K.Ei + K.Ef

1.96 = mghf + ¹/₂mvf²

where;

1.96 = 0.25 x 9.8 x 0.3   +   0.5 x 0.25vf²

1.225 = 0.125vf²

vf² = 1.225/0.125

vf² = 9.8

vf = 3.13 m/s

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The frequency of the 5 waves with a length of 4m hit the shore every 2 seconds is 2.5 Hz.

This is the number of cycles completed by a wave in one second. The s.i

unit of frequency is Hert (Hz).

From the question, to calculate the frequency of 5 waves with length of 4 m that hit the shores every 2 seconds, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The frequency of the wave is 2.5 Hz.

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Answer:

The delicate balance between temperature and magnetic domains is destabilized when a magnet is subjected to high temperatures. If a magnet is exposed to this temperature for an extended length of time or heated over its Curie temperature, it will lose its magnetism and become irreversibly demagnetized.

Explanation:

Answer:25

Explanation: because higher means less kinetic energ

[tex]\\ \rm\rightarrowtail F=-kx[/tex]

[tex]\\ \rm\rightarrowtail F=-20(0.03)[/tex]

[tex]\\ \rm\rightarrowtail F=-0.6N[/tex]

Answer:

See below.

Explanation:

According to the question, we know that,

work done is given by,  [tex]W=qV[/tex]

and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]

therefore equating both the equations we get,

[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]

m= mass of electron =  [tex]9.1*10^{-31} kg[/tex]

q= charge on an electron = [tex]1.6*10^{-19} C[/tex]

v= speed of electron= 700000m/s

substituting the values in the above equation, we get

[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]

(1).  the potential difference that stopped the electron is 1.39 volts.

now the kinetic energy equation is :  2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]

or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]

(2).  the initial kinetic energy of the electron is 1.39eV.

The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.

A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.

This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.

The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.

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Alpha

I hope this helps you

:)

Answer:

C

because urine is waste product