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Vinvika [58]
2 years ago
5

A roller coaster car moving at a velocity of 30 meters/second has a momentum of 2.5 × 104 kilogram meters/second. what is its ma

ss? a. 85 kilograms b. 8.3 × 102 kilograms c. 7.6 × 102kilograms d. 7.6 × 103 kilograms e. 7.6 × 104 kilograms
Physics
1 answer:
frez [133]2 years ago
8 0

The mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.

<h3>How to calculate mass?</h3>

The mass of the roller coaster car can be calculated using the following formula:

P = m × v

Where;

  • P = momentum
  • m = mass
  • v = velocity

m = 2.5 × 10⁴ ÷ 30

m = 8.3 × 10²kg

Therefore, the mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.

Learn more about mass at: brainly.com/question/19694949

#SPJ1

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

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Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

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From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

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\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

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8 0
3 years ago
It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo
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Answer:

L = 1.11 x 10^{6} m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x 10^{5} m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x 10^{-2} m

Length = ?

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we know that,

U = 1/2 C Δv^{2}

U = 52000 J

C = ?

V = 610 volts'

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Plugging in the values into the formula, we get:

0.28 = \frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }

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is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.  

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