It depends, How big is the room. If it was me i would put "High density foam pads" all over the walls in the room, and that would get rid of some sound but put Drywall, or Plasterboard behind the foam pads. That should hopefully get rid of the sound from outside the room, and stop any noise in the background, if your using a mic, for example, a Youtube channel, or Twitch treaming channel. Hope this helps. BYE!!
the friction force provided by the brakes is 30000 N.
<h3>What is friction force?</h3>
Friction force is the force that opposes the motion between two bodies in contact.
To calculate the average friction force provided by the brakes, we apply the formula below.
Formula:
- K.E = F'd............. Equation 1
Where:
- K.E = Kinetic energy of the train
- F' = Friction force provided by the brakes
- d = distance
Make F' the subject of the equation
- F' = K.E/d............ Equation 2
From the question,
Given:
Substitute these values into equation 2
- F' = (8.1 ×10⁶)/270
- F' = 30000 N
Hence, the friction force provided by the brakes is 30000 N
Learn more about friction force here: brainly.com/question/13680415
Explanation:
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Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.
<h3>
Work done</h3>
Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;
W = F × d
Where F is force applied or Weight and d is distance
Also Force = Weight = mass × acceleration due to gravity.
Since gravity is acting on the boxes as it been lift
W = Weight × height from ground level
W = mg × d
Where m is mass of the boxes, g is accelration due to gravity( g = 9.8m/s² ) and d is distance from ground level.
Given the data in the question;
- Since each box has a mass of 7.89 kg
- Mass of the 345 boxes = 345 × 7.89 kg = 2722.05kg
- Distance or height d = 6.0m
To determine the work done, we substitute our values into the expression above.
W = mg × d
W = 2722.05kg × 9.8m/s² × 6.0m
W = 160056.5kgm²/s²
W = 160056.5J
W = 1.6 × 10⁵J
Therefore, Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.
Learn more about work done here: brainly.com/question/26115962
Using the equation, x = (1/2)at^2, the time of the fall is .55s. Using the formula d=rt, we can plug in time and distance to solve for the rate.
Answer: 2.73 m/s