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agasfer [191]
3 years ago
15

Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie

d tangentially to the smaller and larger wheels, respectively. In order to impart the same angular acceleration to each wheel, how large must F2 be? Assume that the hubs and spokes are massless, so that the rotational inertia is given by I = mR2. Answer in units of N.
Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

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By applying the wave equation we know that the displacement on the y-axis is 1.999 micrometers.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

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From the question above, we know that:

y = 2.00cos (15.7x - 858t)

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A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an acceler
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Answer:

(a) m = 33.3 kg

(b) d = 150 m

(c) vf = 30 m/s

Explanation:

Newton's second law to the block:

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a= 3.0 m/s²

(a) Calculating of the  mass of the block:

We replace dta in the formula (1)

F = m*a

100 =  m*3

m = 100 / 3

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Kinematic analysis

Because the block  moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (2)

vf= v₀+a*t   Formula (3)

Where:  

d:displacement in meters (m)  

t : time interval in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

a= 3.0 m/s²

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(b) Distance the block will travel if the force is applied for 10 s

We replace dta in the formula (2):

d= v₀t+ (1/2)*a*t²

d = 0+ (1/2)*(3)*(10)²

d =150 m

(c) Calculate the speed of the block after the force has been applied for 10 s

We replace dta in the formula (3):

vf= v₀+a*t

vf= 0+(3*(10)

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