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PolarNik [594]
2 years ago
8

Bart awoke screaming, shaking, and sweating in his heart was beating rapidly. Homer and Marge, being good parents, had Bart hook

ed up to an EEG. Checking Bart's EEG, they noted that Bart was in stage 4 sleep just before he awoke. From this information, they felt confident in saying that Bart had just experienced __________.
a. a lucid dream
b. an episode of sleep apnea
c. a nightmare
d. sleep terror disorder
Physics
1 answer:
Nuetrik [128]2 years ago
7 0

Answer: D

sleep terror disorder

Explanation:

It is a sleeping disorder causing feelings of panic or dread which typically occur during the first hours of stage 3–4 non-rapid eye movement (NREM) sleep.

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On a topographic map, contour lines create a group of concentric, closed loops. Which of the following features could this indic
posledela

Answer:

b. Hill top

Explanation:

On a topographic map, the closed circles are meant to represent a hill. So if the contour lines are creating a group of concentric closed loops then it must be an indication of a hill.

3 0
3 years ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va
WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V =  3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0
3 years ago
PLS HURRY! `WILL GIVE BRAINLIEST TO FIRST ANSWER
bogdanovich [222]

Answer:

Decrease in speed

Explanation:

Increasing an object's velocity can decrease kinetic energy. *The velocity of an object has no effect on kinetic energy. *Increasing an object's velocity can increase kinetic energy. Increasing an object's velocity can increase kinetic energy.

3 0
3 years ago
Read 2 more answers
A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.27). A light beam of variable wavelength from air i
Rudiy27

Answer:

152.7 nm

Explanation:

Refractive index of glass plate = n = 1.64

Refractive index of oil = n' = 1.27

wavelength = λ = 501 nm = 501 e-9 m

2n t = m  λ , since it undergoes constructive interference.

Thickness = t =  501 × 10⁻⁹ m / 2 (1.64)

                      = 152.7 nm

7 0
3 years ago
A planet similar to the Earth has a radius 5 × 106 m and has an acceleration of gravity of 10 m/s 2 on the planet's surface. The
sp2606 [1]

Answer:

The new period is approximately 3.93 × 10⁻⁷ h

Explanation:

The radius of the planet = 5 × 10⁶ m

The acceleration due to gravity on the planet, a = 10 m/s²

The period of the planet = 25 h

The centripetal force, F_c, is given by the following equation;

F_c = \dfrac{m \cdot v^2}{r}

Where;

v = The linear speed

r = The radius

Therefore, for the apparent weight, W, of an object to be zero, we have;

The weight of the object = The centripetal force of the object

W = Mass, m × Acceleration due to gravity, a

∴ W = F_c

Which gives;

m \times a = \dfrac{m \cdot v^2}{r}

a = \dfrac{v^2}{r}

∵ r = The radius of the planet

We have;

10 = \dfrac{v^2}{5 \times 10^6}

v² = 10 × 5 × 10⁶

v = √(10 × 5 × 10⁶) ≈ 7071.07 m/s

The new frequency = Radius of the planet/(Linear speed component of rotation)

∴ The new frequency = 5 × 10⁶/(7071.07) = 707.107 revolutions per second

The new frequency = 707.107 × 60 × 60 = 2545585.2 revolutions per second

The new period = 1/Frequency ≈ 3.93 × 10⁻⁷ hour.

3 0
3 years ago
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