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IrinaVladis [17]
2 years ago
13

A fancart of mass 0.8 kg initially has a velocity of < 0.9, 0, 0 > m/s. Then the fan is turned on, and the air exerts a co

nstant force of < -0.3, 0, 0 > N on the cart for 1.5 seconds. What is the change in momentum of the fancart over this 1.5 second interval? = < , 0, 0 > kg
Physics
1 answer:
Dmitry [639]2 years ago
6 0

Answer:

<-0.45, 0, 0> kgm/s

Explanation:

The change in momentum would equal to the impulse generated by force F = <-0.3, 0, 0> over time duration of Δt = 1.5 second. The impulse is defined as the product of force F and the duration Δt

\Delta p = F \Delta t = (-0.3) * 1.5 = -0.45 kgm/s

So the change in momentum is <-0.45, 0, 0> kgm/s

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Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the
makvit [3.9K]

Answer:

\mu_k=0.101

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

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When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

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\mu_k=\dfrac{kx^2}{2mgh}

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3 0
3 years ago
A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>​
Vadim26 [7]

Answer:

\huge\boxed{\sf f=0.55 \ Hz}

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }

<u>Solution:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}

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2 years ago
An electric motor converts electrical energy into (blank) energy
Delicious77 [7]

Answer:

Mechanical

Explanation:

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2 years ago
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