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3 years ago

14

An electrician timed a box traveling over a conveyor and measured 25 sec to travel 30 feet. If the gearbox ratio is 20:1 and the

driver drum is 24", what is the RPM of the motor?

1 answer:

nataly862011 [7]3 years ago

4 0

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At which position is the northern hemisphere experiencing winter?

Leto [7]

The answer would be point A.

Hope this helped you.

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3 years ago

Sam drives her scooter 7 kilometres north. She stops for lunch and then drives

Talja [164]

Answer:

1. Distance travelled = 12 km.

2. Displacement = 8.6 km

Explanation:

From the question given above, the following data were obtained:

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance =?

Displacement =?

1. Determination of the distance travelled.

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance (dₜ) =?

dₜ = d₁ + d₂

dₜ = 7 + 5

dₜ = 12 km

2. Determination of the displacement.

In the attached photo, R is the displacement.

We can obtain the value of R by using the pythagoras theory as illustrated below:

R² = 7² + 5²

R² = 49 + 25

R² = 74

Take the square root of both side

R = √74

R = 8.6 km

8 0

2 years ago

Power is measured in unit of Joules per second or

Rasek [7]

D watts yw .yea yww

5 0

2 years ago

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

B)

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

2 years ago

What is the force felt by the 64-kg occupant of the car? Express your answer to two significant figures and include the appropri

Yakvenalex [24]

Answer:

The force is $-1.67\times10^{5}\ N$

Explanation:

Given that,

Mass of car = 64 kg

Suppose, a 1400-kg car that stops from 34 km/h on a distance of 1.7 cm.

We need to calculate the acceleration

Using formula of acceleration

$v^2-u^2=2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the formula

$0^2-(34\times\dfrac{5}{18})^2=2\times a\times 1.7\times10^{-2}$

$a=\dfrac{(34\times\dfrac{5}{18})^2}{2\times1.7\times10^{-2}}$

$a=-2623.45\ m/s²$

We need to calculate the force

Using formula of force

$F=ma$

$F=64\times(-2623.45)$

$F=-1.67\times10^{5}\ N$

Negative sign shows the direction of the force is in the direction opposite to the initial velocity.

Hence, The force is $-1.67\times10^{5}\ N$

5 0

2 years ago