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n200080 [17]
3 years ago
12

Un auto de 2,300 kg está estacionado sobre una rampa inclinada 26.0°. Obtenga la tensión en la cadena.

Physics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

T = mgsinθ = 2300(9.8)sin26.0 = 9880.88 ≈ 9900 N

Explanation:

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9)A 64 kg parent and a 16 kg child meet at the center of an ice rink. They place their hands together and push. (A) Is the
Colt1911 [192]

Answer:

(A) <u>The same</u>

(B) <u>More</u>

(C) The magnitude of the parent's acceleration is 0.625 m/s²

Explanation:

The given parameters are;

The mass of the parent, m₁ = 64 kg

The mass of the child, m₂ = 16 kg

(A) By Newton's third law of motion, action and reaction are equal and opposite

Therefore, the action of the parent on the child is equal to the reaction of the child on the parent and vice versa

Therefore, the force experienced by the child is <u>the same</u> as the force experienced by the parent

(B) Newton's second law states that an objects acceleration is directly proportional to the applied force and inversely proportional to the mass of the object

Therefore, the parent and the child both experience the same force but the mass of the child is less than the mass of the parent and therefore, by Newton's second law, the acceleration of the child will be <u>more</u> than the acceleration of the parent for the same given force

(C) The acceleration of the child, a₂ = 2.5 m/s²

Let F₁ represent the force experienced by the parent, let F₂ represent the force experienced by the child and let a₁ represent the magnitude of the parent's acceleration

By Newton's third law, we have;

F₁ = F₂

Force, F = Mass, m × Acceleration, a

We can write, F = m × a

Therefore;

F₁ = m₁ × a₁ and F₂ = m₂ × a₂

∴ F₁ = F₂ gives;

m₁ × a₁ = m₂ × a₂

a₁ = (m₂ × a₂)/m₁ = (16 × 2.5)/64 = 0.625

∴ The magnitude of the parent's acceleration = a₁ = 0.625 m/s²

5 0
3 years ago
An alternating-current (AC) source supplies a sinusoidally varying voltage that can be described with the function v of t is equ
Marrrta [24]

Answer:

ω, the angular frequency of the source equals 377 rad/s

Explanation:

From the question, V(t) = V cosωt.

Now, ω = the angular frequency of the sinusoidal wave is given by

ω = 2πf where f = the frequency of the source = 60 Hz

So, the angular frequency of the source ,ω = 2π × the frequency of the source.

So, ω = 2πf

ω = 2π × 60 Hz

ω = 120π rad/s

ω = 376.99 rad/s

ω ≅ 377 rad/s

So, ω, the angular frequency of the source equals 377 rad/s

3 0
3 years ago
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