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Ede4ka [16]
3 years ago
7

What is the correct answer?

Physics
1 answer:
sashaice [31]3 years ago
7 0

Answer:

D) 4

Explanation:

Roots of a polynomial must be factors of the last term.

In this case, the factors of 6 are +1, -1, +2, -2, +3, -3, +6, -6. The only factor that doesn't show up, given the options, is 4. This means that D is the correct answer.

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You did 200 joules of work lifting a 150-newton backpack. How high did you lift the backpack?
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We know that:
W=Fs
200J=150N*s
s=200J/150N
s=1,33m
8 0
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If someone is moving at a constant speed of 100 km/h and accelerate to 120km /h in 10 s what is the acceleration
Alex Ar [27]

The acceleration should be a gain of 2 km/h per second

5 0
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When the volcano Krakatoa erupted in 1883, it was heard 5000 km away. Which statement about the sound from the volcano is not co
lina2011 [118]
Answer a) is incorrect as sound does not travel in a vacuum.
7 0
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The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the
Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t      ⇒     α = (ωfin - ωin) / t

⇒     α = (0 rev/min - 113 rev/min) / (60 min)

⇒     α = - 1.883 rev/min²

6 0
3 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
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