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2 years ago

6

Major difference between cross belt and open belt drives

1 answer:

pav-90 [236]2 years ago

8 0

Answer:

The belt in an open belt drive travels from one pulley's top to the top of another without intersecting. In cross belt drive, the belt crosses itself by moving from the top of one pulley to the bottom of another. Every revolution, the entire belt remains in the same plane.

Hope this helps!!!

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1. ELECTRICAL SHOCK

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14. E
15. N

3 0

4 years ago

Two vehicles arrive at an uncontrolled intersection from different streets at the same time 1.The driver on the right must yield

ss7ja [257]

3. Both vehicles must stop

4 0

3 years ago

Which of these cars traveled faster during time interval <br> please show solution

Mazyrski [523]

Answer:

I think D is correct

Explanation:

C is decreasing function, probably worst

A is arctan -> in radian, the rate of increasing is very slow-> second worst

B(14) = ln(9*14) = 4.8

D(14) = sqrt(8+14^2)=14.2

3 0

3 years ago

If 5000 N of thrust is acting to the left, and 4300 N of drag is acting to the right, what is the magnitude and direction of the

Kipish [7]

Answer:

700 N acting to the left.

8 0

3 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago