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sergejj [24]
2 years ago
10

Two construction cranes are each able to lift a maximum load of 20000 N to a height of 250 m. However, one crane can lift that l

oad in 1 6 the time it takes the other. How much more power does the faster crane have?
Physics
1 answer:
Neporo4naja [7]2 years ago
5 0

Answer:

Explanation:

Given

Load W=20000\ N

height to which load is raised h=250\ m

Another crane take \frac{1}{6} th time to lift the load

Energy required required to lift the Weight

E=W\times h

E=20000\times 250

E=5,000,000\ J

Suppose P_1 and P_2 is the Power required to lift the weight in t and \frac{t}{6} time

E=P_1\times t

E=P_2\times \frac{t}{6}

P_1\times t=P_2\times \frac{t}{6}

thus

P_2=6P_1

Second Crane requires 6 times  more power than the slow crane                                              

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One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
A brick of mass 2.0kg is at rest. It falls to the ground through a
Lisa [10]

Answer:

I may not have the answer so i'll just give up some hints.

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

I hope my little bit (big you may say) hint help you with your question.

5 0
2 years ago
Someone please help!!!
aivan3 [116]

Answer:

A

Explanation:

hope this helps!

6 0
2 years ago
Water moves through a constricted pipe in steady, ideal flow. At the
Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: 1.8\cdot 10^{-3} m^3/s

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2

where

p_1=1.75\cdot 10^4 Pa is the pressure in the lower section of the tube

h_1 = 0 is the heigth of the lower section

\rho=1000 kg/m^3 is the density of water

g=9.8 m/s^2 is the acceleration of gravity

v_1 is the speed of the water in the lower pipe

p_2 is the pressure in the higher section

h_2 = 0.250 m is the height in the higher pipe

v_2 is hte speed in the higher section

We can re-write the equation as

v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho} (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-section of the lower pipe, with

r_1 = 3.00 cm =0.03 m is the radius of the lower pipe (half the diameter)

A_2 = \pi r_2^2 is the cross-section of the higher pipe, with

r_2 = 1.50 cm = 0.015 m (radius of the higher pipe)

So we get

r_1^2 v_1 = r_2^2 v_2

And so

v_2 = \frac{r_1^2}{r_2^2}v_1 (2)

Substituting into (1), we find the speed in the lower section:

v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s

B)

Now we can use equation (2) to find the speed in the lower section:

v_2 = \frac{r_1^2}{r_2^2}v_1

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s

C)

The volume flow rate of the water passing through the pipe is given by

V=Av

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume  flow rate is constant, so

r_1=0.03 cm

v_1=0.638 m/s

Therefore, the volume flow rate is

V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0
2 years ago
A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s, and the stone is 1.50 m above the ground when lau
arlik [135]

Answer: a) 19.21m b) 3.92secs

Explanation:

a) Maximum height reached by the object is the height reached by an object before falling freely under gravity.

Maximum height = U²/2g

U is the initial velocity = 19.6m/s

g is acceleration due to gravity = 10m/s²

Maximum Height = 19.6²/2(10)

H = 19.21m

b) The time elapsed before the stone hits the ground is the time of flight T= 2U/g

T= 2(19.6)/10

T = 39.2/10

Time elapsed is 3.92secs

5 0
3 years ago
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