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marissa [1.9K]
1 year ago
7

A gyroscope flywheel of radius 1.96 cm is accelerated from rest at 13.0 rad/s2 until its angular speed is 2270 rev/min. (a) What

is the tangential acceleration of a point on the rim of the flywheel during this spin-up process
Physics
1 answer:
nika2105 [10]1 year ago
5 0

Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².

Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.

Given,

Radius of flywheel (r) = 1.96 cm = 0.0196m

Angular acceleration (α)= 13.0 rad/s²

The tangential acceleration formula is at=rα

where, α is the angular acceleration, and r is the radius of the circle.

using the formula; at=rα  = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².

The tangential acceleration is 0.2548 m/s².

Learn more about the Tangential acceleration with the help of the following link:

brainly.com/question/15743294

#SPJ4

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What is the frequency of a wave whose wavelength is 0.7m and whose velocity is 120m/s?
ANEK [815]

Answer:

f=171.43Hz

Explanation:

Wave frequency is the number of waves that pass a fixed point in a given amount of time.

The frequency formula is: f=v÷λ, where <em>v</em> is the velocity and <em>λ</em> is the wavelength.

Then replacing with the data of the problem,

f=\frac{120\frac{m}{s} }{0.7m}

f=171.43\frac{1}{s}

f=171.43 Hz (because \frac{1}{s} = Hz, 1 hertz equals 1 wave passing a fixed point in 1 second).

5 0
3 years ago
The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How
sammy [17]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

6 0
3 years ago
A hockey puck slides 55.0 m along the length of the rink in just 1.25 s. The slight friction between the puck and the ice provid
topjm [15]

Answer:

44.8 m/s

Explanation:

Use the Initial Speed Formula:

InS = 2(d/t) - Final Speed

InS = 2(55/1,25) - 43.2

InS = 2.44 - 43,2

InS = 88 - 43,2

InS = 44.8 m/s

3 0
2 years ago
Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains
tester [92]

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v =  (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

3 0
3 years ago
50 POINTS‼️‼️‼️‼️‼️
Inessa05 [86]

Answer:

Explanation:

the yellow box is -5

3 0
3 years ago
Read 2 more answers
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