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nevsk [136]
3 years ago
6

I shared a picture of the problem. It’s a basic Physics question and an Algebra question.

Physics
1 answer:
julia-pushkina [17]3 years ago
3 0

Hence the expression of ω in terms of m and k is

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

Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

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

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Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
3 years ago
when you park alongside a curb, the front and back wheels must be parallel and no more than ....... inches from the curb
Rainbow [258]

Between 12 to 18 inches.

The front and back wheels must be parallel and within at least 12 to 18 inches of the curb while parking next to a curb on a level street. If there is no curb, park parallel to the road.

➤ The following specific parking guidelines apply to painted colored curbs:

Only stop at a white light long enough to pick up or drop off mail or passengers.

Green-Park only for a short while. For time limits, look for a sign that is put adjacent to the green zone or the time limit that is painted on the curb.

Yellow-Stop for no longer than the indicated amount of time to load or unload cargo or passengers. Noncommercial vehicle drivers are typically expected to stay behind the wheel.

No stopping, standing, or parking is permitted (buses may stop at a red zone marked for buses).

Blue-parking is only allowed for a disabled person or the driver of a disabled person who has a placard or special license plate for disabled people or disabled veterans. A no-parking zone is defined as a crosshatched (diagonal lines) area adjacent to a designated disabled parking space.

Find more on parking related questions at : brainly.in/question/7229826

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7 0
1 year ago
The current perception of the arrangement of an atom is represented in the picture at right.
yarga [219]

Answer:

theroy

Explanation:

6 0
3 years ago
A spring has a spring constant of 90N/m.How much potential energy does it store when stretched by 2 cm?
il63 [147K]

Answer:

The potential energy stored in the spring is 0.018 J.

Explanation:

Given;

spring constant, k = 90 N/m

extension of the spring, x = 2 cm = 0.02 m

The potential energy stored in the spring is calculated as;

U = ¹/₂kx²

where;

U is the potential energy stored in the spring

Substitute the given values in the equation above;

U = ¹/₂  x  90 N/m  x  (0.02 m)²

U = 0.018 J

Therefore, the  potential energy stored in the spring is 0.018 J.

5 0
3 years ago
An object of mass 3.4 kg is moving in a straight line with kinetic energy 59.177 J. A force is applied in the direction of its m
rusak2 [61]

Answer:

Its momentum is multiplied by a factor of 1.25

Explanation:

First, we <u>calculate the initial velocity of the object</u>:

  • K = 0.5 * m * v₁²
  • 59.177 J = 0.5 * 3.4 kg * v₁²
  • v₁ = 5.9 m/s

With that velocity we can <u>calculate the initial momentum of the object</u>:

  • p₁ = v₁ * m
  • p₁ = 20.06 kg·m/s

Then we <u>calculate the velocity of the object once its kinetic energy has increased</u>:

  • (59.177 J) * 1.57 = 0.5 * 3.4 kg * v₂²
  • v₂ = 7.4 m/s

And <u>calculate the second momentum of the object</u>:

  • p₂ = v₂ * m
  • p₂ = 25.16 kg·m/s

Finally we <u>calculate the factor</u>:

  • p₂/p₁ = 1.25
3 0
2 years ago
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