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1 year ago

Good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.

Physics

2 answers:

Paul [167]1 year ago

6 0

Drinking-Water - Direct

Eating Habits - Direct

Washing Clothes - Indirect

Gekata [30.6K]1 year ago

4 0

Drinking Water - Direct
Our Eating Habits - Direct
Washing Clothes - Indirect

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Technician A says that a proper U-joint inspection can be done with the driveshaft in place on the vehicle. Technician B says th

Julli [10]

Answer:

TECHNICIAN B

Explanation: Drive shaft is an elongated part of the drive system made of steel of a vehicle that transmits the ENGINE TORQUE to the WHEELS of a vehicle. For proper inspection a technician is expected to completely remove the driveshaft to check all its compactment. It is Cylindrical and usually used for REAR-WHEEL-VEHICLES. Drive shafts are known by different names like propeller shaft,Cardan shaft etc it has different types which range from

One-piece drive shaft ,Two-piece drive shaft to Slip-in-tube drive shaft .

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2 years ago

Let us remember your previous lesson on Physical Press Components! Directions: Analyze the folawna fitness components. Put a che

mart [117]

Answer:

5 Muscular and strength

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2 years ago

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

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2 years ago

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Why can the atmosphere hold on to heat

Naddika [18.5K]

Answer:

A. Air

Explanation:

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2 years ago

A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal

givi [52]

I'm gonna have to assume the girl is on the right side and boy on left.
The net force is the sum of all forces on an object (includes negatives).
Let's say the force of the boy is variable b. Use the formula F = ma.

b + 3.5 = 0.2(2.5)

This is now simple algebra. Solve to get that the boty is exerting a force of -3N to the left.

5 0

2 years ago

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