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saw5 [17]
3 years ago
9

Which is not true of the weak nuclear force​

Physics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

options?

Explanation:

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The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'
Svetlanka [38]

Answer:

79

Explanation:

4 0
2 years ago
Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri
igor_vitrenko [27]

Answer:

a) the particles are <em>0.217 m </em>apart

b) <em>the particles are moving in the same direction</em>.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

    = -0.155 A + 0.372 A

    = 0.217 A

since A = 1 m

Thus,

<em>Δx  = 0.217 m</em>

<em></em>

<em></em>

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

   = (-πA / T) sin(2πt / T)

   = -(π(1) / 1.5) sin(2π(0.45) / 1.5)

   = -1.99

and,

v₂ = dx₂ / dt

   = (-πA / T) sin((2πt / T) + π/6)

   = -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

   = -1.40

Since both v₁ and v₂ are negative, this shows that <em>the particles are moving in the same direction</em>.

6 0
3 years ago
A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,
Andreyy89

Answer:

-450 m/s

Explanation:

Momentum is conserved.

p₀ = p

0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v

v = -450 m/s

6 0
3 years ago
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
A parsec is a measurement of: magnitude distance position speed
guajiro [1.7K]
A parsec is a measurement of distance.
6 0
3 years ago
Read 2 more answers
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