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3 years ago

A delusion is best characterized as

Physics

2 answers:

deff fn [24]3 years ago

7 0

A disorder where a person loses contact with the real world, and experiences unreasonable ideas.

balu736 [363]3 years ago

5 0

Answer:

This is the most common form of delusional disorder. In this form, the affected person fears they are being stalked, spied upon, obstructed, poisoned, conspired against or harassed by other individuals or an organization.

Explanation:

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What kind of waves can travel through empty space or a vacuum?

satela [25.4K]

Answer:

the answer is B.

4 0

4 years ago

Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p

Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0

4 years ago

Particles q1, q2, and q3 are in a straight line. Particles q1 = -28.1 μC, q2 = +25.5 μC, and q3 = -47.9 μC. Particles q1 and q2

Arturiano [62]

Answer:-88.4

Explanation:

3 0

3 years ago

Read 2 more answers

The distance from earth to a NASA satellite traveling through the solar system is 6.0 × 109m. A command is sent to the satellite

alexandr1967 [171]

Answer:

the signal it takes 40 s to get the answer

Explanation:

As the satellite is in space the speed of radiation (waves9 is the speed of light in a vacuum, which is constant 3. 108 m / s, so we can use the uniform motion ratios to find the time.

v = d / t

t = d / v

t = 6.0 109 / 3. 108

t = 2 101 s

t = 20 s

This is the time it takes to get the signal from the earth to the satellite and it takes the same time to return, so since they send the signal it takes 40 s to get the answer

5 0

4 years ago

The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

Elanso [62]

On Earth, the period of a pendulum is given by:
$T_{earth}=2\pi \sqrt{ \frac{L}{g_{earth} }$
where L is the length of the pendulum and $g_{earth}=9.81~m/s^2$ is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
$T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }$
where $g_{mars}=3.71~m/s^2$ is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
$\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} } = \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63$
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.

6 0

4 years ago