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anastassius [24]
3 years ago
12

i cant figure out how to delete a question on this website. so i am writing this until i figure it out.

Physics
1 answer:
sergij07 [2.7K]3 years ago
5 0

Answer:

I'm guessing it would remain a mystery to us all

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How many nanometers is 0.0357 m?
tangare [24]

Answer:

35700000

Explanation:

multiply the meter value by 1e+9

3 0
4 years ago
Read 2 more answers
A bug of mass 2.2 g is sitting at the edge of a cd of radius 3.0 cm. if the cd is spinning at 280 rpm, what is the angular momen
m_a_m_a [10]
M = 2.2 g = 2.2 x 10⁻³ kg, the mass of the bug.
r = 3.0 cm = 0.03 m, the radial distance from the center.

The angular speed is
ω = 280 rpm
    = (280 rev/min)*(2π rad/rev)*(1/60 min/s)
    = 29.3215 rad/s

The moment of inertia of the bug is
I = mr²
  = (2.2 x 10⁻³ kg)*(0.03 m)²
  = 1.98 x 10⁻⁶ kg-m²

Calculate the angular momentum of the bug.
J = Iω
  = (1.98 x 10⁻⁶ kg-m²)*(29.3215 rad/s)
  = 5.806 x 10⁻⁵ (kg-m²)/s

Answer: 5.806 x 10⁻⁵ (kg-m²)/s

5 0
4 years ago
A car with the mass of '800 kg is moving at a velocity of 34 m/s. What is 10 points
klio [65]

Answer:462400joules

Explanation:

Mass=m=800kg velocity=v=34m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(800x34^2)/2

Kinetic energy =(800x34x34)/2

kinetic energy =924800/2

Kinetic energy =462400joules

7 0
3 years ago
I need some help. This is due today. Please
Law Incorporation [45]
I need a better picture
7 0
4 years ago
A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t
Marianna [84]

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = \frac{a}{g}

μ = \frac{1.25}{9.8}

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

6 0
4 years ago
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