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NeTakaya
3 years ago
15

At how many rpm must the disk turn to raise the elevator at 35.0 cm/s

Physics
1 answer:
k0ka [10]3 years ago
3 0
<span>Let R be the radius of the disk, in centimeters. 35.0 cm/s = 35.0 * 60 cm/min = 2100 cm/min. One revolution of the disk is 2*pi*R, in cm. So the number of revolutions per minute is 2100 / (2*pi*R) = 1050/(pi*R), approximately 334/R.</span>
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The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris
Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

        \sum F_{x} = 0

        T + F_{AB} Sin 60 = 0 ...... (1)

       \sum F_{y} = 0

       F_{AB} Cos 60 + W = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

               \sigma_{allow} = \frac{T}{A}

       T = (20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}

          = 3925 kip

From equation (1),   F_{AB}Sin (60^{o})  = -3925

               F_{AB} \times -0.304 8 = -3925

             F_{AB} = 12877.29 kip

From equation (2),    -12877.29 (Cos 60) + W = 0

         -12877.29 kip \times \frac{1}{2} + W = 0

                           W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0
3 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
3 years ago
An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-dia
11111nata11111 [884]

Answer:

1.4E-3J

Explanation:

Given that

Time = 8hrs = 28.8E3 seconds

Intensity= 90dB

D= 0.008m

Radius= 0.004m

So intensity is sound level Bis

10dBlog(I/Io)

I= 10 (B/10dB)Io

= 10( 90/10) x 10^-12

=0.001W/m²

But we know that

I = P/A

P= I πr²

= 5.02 x10^-8W

But energy is power x time

So E= 5.02E-8 x 28.8E3

= 1.4E-3J

5 0
3 years ago
Read 2 more answers
An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,
Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0
3 years ago
5. Stopping a fast-moving object is harder than stopping a slow-moving<br> one.<br> True<br> False
Strike441 [17]
True because well it’s moving fast lol sometimes ur eyes have a hard time following its speed
6 0
3 years ago
Read 2 more answers
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