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juin [17]
3 years ago
6

A grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 12

rad/s over the next 4.0 seconds. Assume that the angular acceleration is constant.What is the magnitude of the angular acceleration of the grindstone?
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

1 rad/s^2

Explanation:

The magnitude of the constant angular acceleration is the change in angular velocity over a unit of time. So if the angular speed increases from 8 rad/s to 12 rad/s in 4 seconds then the angular acceleration is:

\alpha = \frac{\Delta \omega}{\Delta t} = \frac{12 - 8}{4} = \frac{4}{4} = 1 rad/s^2

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astraxan [27]
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= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km 
The work done by the aircraft overcoming the wind is 
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N 
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6 0
3 years ago
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Andrew [12]
The last one, the soil will become weak & unable to support plant growth
4 0
3 years ago
A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e
IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm  - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0
3 years ago
an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?
gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

F_{net} = F_g - F = m\cdot g - F\\

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N

The tension force F in the supporting cables is 7245N


3 0
3 years ago
A ball is thrown into the air with a vertical velocity of 50 m/s and a horizontal
daser333 [38]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

we know that u*sin(theta) = vertical velocity

so the time taken to reach the maximum height or the time of Ascent is equal to

T = Usin(theta) ÷ g, here g = 9.8 m/s^2

so we get as,

T = 50/9.8

T = 5.10 seconds

thus the time taken to reach max height is 5.10 seconds.

5 0
3 years ago
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