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3 years ago

5

Two wheel gears are connected by a chain. The larger gear has a radius of 5 centimeters and the

1 answer:

DENIUS [597]3 years ago

6 0

Answer:

The center velocity of both gears is same & equal to 2034.72 $\frac{cm}{min}$

Step-by-step explanation:

Given data

$R_1$ = 5 cm

$R_2$ = 3 cm

$N_2 =$ 36 rev in 20 sec

$\omega_2 = 2 \pi N_2$

$\omega_2$ = 2 × 3.14 × 1.8

$\omega_2$ = 11.304 $\frac{rad}{s}$

We know that for the set of gears

$R_1\omega_1 = R_2\omega_2$

$5 \omega_1 =$ 3 × 11.304

$\omega_1$ = 6.7824 $\frac{rad}{s}$

liner velocity of the first gear

$V_1 = R_1 \omega_1$

$V_1 =$ 5 × 6.7824

$V_1 =$ 33.912 $\frac{cm}{s}$

$V_1 =$ 20.34.72 $\frac{cm}{min}$

liner velocity of the second gear

$V_{2} = R_2 \omega_2$

$V_2 =$ 3 × 11.304

$V_2 =$ 2034.72 $\frac{cm}{min}$

Therefore the center velocity of both gears is same & equal to 2034.72 $\frac{cm}{min}$

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3 years ago

Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) ∈ R if and only if ad = bc. Ar

Veseljchak [2.6K]

Answer:

The given relation R is equivalence relation.

Step-by-step explanation:

Given that:

$((a, b), (c, d))\in R$

Where $R$ is the relation on the set of ordered pairs of positive integers.

To prove, a relation R to be equivalence relation we need to prove that the relation is reflexive, symmetric and transitive.

1. First of all, let us check reflexive property:

Reflexive property means:

$\forall a \in A \Rightarrow (a,a) \in R$

Here we need to prove:

$\forall (a, b) \in A \Rightarrow ((a,b), (a,b)) \in R$

As per the given relation:

$((a,b), (a,b) ) \Rightarrow ab =ab$ which is true.

$\therefore$ R is reflexive.

2. Now, let us check symmetric property:

Symmetric property means:

$\forall \{a,b\} \in A\ if\ (a,b) \in R \Rightarrow (b,a) \in R$

Here we need to prove:

$\forall {(a, b),(c,d)} \in A \ if\ ((a,b),(c,d)) \in R \Rightarrow ((c,d),(a,b)) \in R$

As per the given relation:

$((a,b),(c,d)) \in R$ means $ad = bc$

$((c,d),(a,b)) \in R$ means $cb = da\ or\ ad =bc$

Hence true.

$\therefore$ R is symmetric.

3. R to be transitive, we need to prove:

$if ((a,b),(c,d)),((c,d),(e,f)) \in R \Rightarrow ((a,b),(e,f)) \in R$

$((a,b),(c,d)) \in R$ means $ad = cb$ .... (1)

$((c,d), (e,f)) \in R$ means $fc = ed$ ...... (2)

To prove:

To be $((a,b), (e,f)) \in R$ we need to prove: $fa = be$

Multiply (1) with (2):

$adcf = bcde\\\Rightarrow fa = be$

So, R is transitive as well.

Hence proved that R is an equivalence relation.

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3 years ago

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dexar [7]

Answer:

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Step-by-step explanation:

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