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Alik [6]
4 years ago
5

A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 18 rpm,

in 13.0 s. Assume the merry-go-round is a uniform disk of radius 2.7 m; and has a mass of 800 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque. What force is required at the edge?
Physics
1 answer:
puteri [66]4 years ago
4 0

Answer:

F = 176.175 N

Explanation:

given,  

radius of merry - go - round = 2.7 m

mass of the disk = 800 kg  

speed of the merry- go-round = 18 rpm  

                                                 = 18 \dfrac{2\pi }{60}

                                                 = 1.88 rad/s

time = 13 s

mass of two children = 25 kg  

ω = ω₀ + α t

1.88 = 0 + α(13)

α = 0.145 rad/s²

we know

τ = I α

so,

I = \dfrac{1}{2}MR^2+ 2 MR^2  

I = \dfrac{1}{2}\times 800 \times 2.7^2+ 2\times 25\times 2.7^2

I = 3280.5 kg.m^2  

τ = I α

τ = 3280.5 x 0.145

τ = 475.67 N m

τ = F x r

475.67 = F x 2.7

F = 176.175 N

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A = 40
mihalych1998 [28]

Answer:

A DAY IN THE LIFE OF METIS JOURNAL, I WROTE IT HOPE U LIKE IT ;))) XD

We, who live in the west, were Nomadic. So we always moved from place to place to hunt buffalo and other fur-bearing animals. And now again we have to shift everything we collected , like, all the essential needs we collected. It’s really hard to shift stuff, But when you move to a new place you get to see new beautiful places. They are just so satisfying to see! It’s like all my work(like the shifting work) is paid off. I love it! We always meet the Hudson’s bay company employees, and give them furs which we took from the bissons and other furry animals we hunted. And in exchange they give us their (Hudson’s Bay company) goods. Which is good! Because then we or I won't be able to enjoy that delicious pemmican!

 The one reason that makes me proud of my parents is that because of them I am like a translator for both sides in the fur trade! Anyways coming back to the present. As I said before, we were moving today to another place, But on the way some people along with my father will ride the famous York boats carrying supplies to fur posts. I begged my father to take me with him! And as usual he gave me some silly reason for me to stay. And now since I don’t have anything to do I will-..Nevermind, My mom’s calling me. Okay after helping my mom keep the fur posts into the York boats me , my mom, and the others are just relaxing and enjoying nature just to take a break. Since, I am so obsessed with my Journal notebook today I’m just going to sit here and right what comes to my mind. I’m not like the others who always seem to be talking with someone. I like my own space! 

I kind of miss my dad already, but he and his fellow members need to travel the waterways from the red river further west and north to hudson bay. These red river carts also travel the trek back and forth over land to St. Paul, Minnesota and to posts in saskatchewan.  Speaking about the red river, the only large settlement in the region is the red colony. About 12000 people lived in this colony around Fort Garry. Some were the original Selkirk settlers or their descendents. They had come from Scotland with lord Selkirk when he established the colony in 1811. Many of these original settlers were poor farmers who had been displaced  from their lands in scotland. They had endured many hardships in the early years of settlement, including foods, hunger, and sickness. I love studying about other colonies, cultures , history etc.

Explanation:

write this as your answer you will get extra marks

4 0
3 years ago
The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

distance ( d ) = 11.6 meter                                       ( m = mass of the car )

Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0
3 years ago
Hii:) I need help with qn 1 & please explain if possible too :) , thanks!
andrezito [222]

1. In the first 1.5 seconds, the lift accelerates from 0 to 3m/s. By definition, the acceleration is the ratio between the change in velocity and the time elapsed to change the velocity.

The change in velocity is \Delta v = v_{\text{final}}-v_{\text{initial}}=3-0=3.

The time elapsed is 1.5 seconds, so the acceleration is

a = \dfrac{3}{1.5}=2

meters per second squared.

2. We know, from the previous point, that the lift travelled 20m from the first floor. Since it returns to the first floor after the ascent, it must travel again those same 20m, just in reverse (descending instead of ascending). So, the total distance travelled is 20+20=40 meters.

The displacement, though, is zero, because it measures the distance between the starting and ending point of a certain motion. Since the lift starts and ends its motion at the same place (the first floor), its total displacement is zero.

3 0
3 years ago
Which is a negatively charged particle in an atom?
koban [17]

Answer:

electrons

Explanation:

electrons are negatively charged particles in an atom

3 0
3 years ago
Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.
lapo4ka [179]

Answer:

This is how I do it:

  • 5.24 rounded to one significant figure is 5
  • 43.261 rounded to one significant figure is 40.
  • 7.3458 rounded to one significant figure is 7.
  • 5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0
3 years ago
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