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Bad White [126]
3 years ago
15

TRUE or FALSE: Feeling "weightless" can be the result of accelerating downwards towards Earth.

Physics
1 answer:
goblinko [34]3 years ago
4 0
I believe it would be false because in outer space, everything is weightless due to the weak gravitational force which makes things float in the air. Weightless means no weight to it and if we’re accelerating down that means the gravitational is working but in this case it would probably be false
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A particle that has mass m and charge q enters a uniform magnetic field that has magnitude B and is directed along the x axis. T
Usimov [2.4K]

Answer:

a) The trajectory will be a helical path.

b) θ = 2*π rad

Explanation:

a) Since the initial velocity of the particle has a component parallel (x-component) to the magnetic  field B , the trajectory will be a helical path.

b) Given

t = 2*π*m/(q*B)

We can use the equation

θ = ω*Δt

where

θ is the angular displacement

ω is the angular speed, which is obtained as follows:

ω = q*B/m

then we have

θ = (q*B/m)*2*π*m/(q*B)

⇒  θ = 2*π rad

6 0
3 years ago
Particles in which state of matter are the most likely to interact with each other to cause a chemical reaction?
Mazyrski [523]
I think when particles are in the gas form they are most likely to cause a chemical reaction considering the fact that gas particles aren't very controllable and it would make sense that they would react unexpectedly
7 0
2 years ago
A bionic man running at 6.5 m/sec , east is acceleration at a uniform rate of 1.5 m/sec^2 east over a displacement of 100.0 m ea
tangare [24]
Given:
u = 6.5 m/s, initial velocity 
a = 1.5 m/s², acceleration
s = 100.0 m, displacement

Let v =  the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s

Answer: 18.5 m/s
5 0
3 years ago
How do you write the scientific name delis cactus correctly
Talja [164]
Delis cactus is the correct way to write it.
8 0
3 years ago
Read 2 more answers
To make a given sound seem twice as loud, how should a musician change the intensity of the sound?
Serhud [2]

Answer:

C. Quadruple the intensity

Explanation:

The intensity of the sound is proportional to square of amplitude of the sound.

I ∝ A²

\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.

A₂ = 2A₁

I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity

3 0
3 years ago
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