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N76 [4]
3 years ago
12

An elevator does 9.75 x 10(4) J of work on a person riding up to another floor. How much energy does the person gain

Physics
1 answer:
Serjik [45]3 years ago
5 0

Answer:

9.75 x 10^4 J

Explanation:

Work done, W = 9.75 x 10^4 J

According to the work energy theorem, the change in kinetic energy is equal to the work done by all the forces.

So, here work done is 9.75 x 10^4 J so the change in kinetic energy is 9.75 x 10^4 J.

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Can you help with this question please
Iteru [2.4K]

Answer:

First answer to the first question is Two people pulling a rope with the same force in a opposite direction.  The other one would be 2.72N        

Explanation:

Hope this helps you :)

7 0
2 years ago
Atoms in Group 18 elements are inert (chemically unreactive) because ___________________________. A they combined to form molecu
Rom4ik [11]

When you say full valence shell, are you talking about a valence electron shell?

I am learning about atoms and i know a little bit

6 0
2 years ago
Read 2 more answers
A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg
ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s

Therefore, the final components of velocity of the composite object is 3.33 m/s.

8 0
2 years ago
Dans un tube en U contenant du mercure ,on verse de l'autre côté de l'acide sulfurique de densité 1,84 et de l'autre côté de l'a
Iteru [2.4K]

Explanation:

Unclear question. The clear rendering reads;

"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?

6 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
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