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lesantik [10]
3 years ago
5

Someone please please help me with this.

Physics
1 answer:
Softa [21]3 years ago
4 0
600 g=.6 kg
mgh=pe
.6(10)(1)=6
therefore, the basketball will have roughly 6 joules of ke.
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3 years ago
A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T
Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= \frac{1}{2}kx² = \frac{1}{2}(530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mgh_o = \frac{1}{2}kx² - mgx

And, h_o= ( \frac{1}{2}kx² - mgx )/(mg) = [\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)]/(0.645x9.8)    

   h_o=   0.78m            

d) Now, if the initial initial height of block is 3h_o

h_o = 3 x 0.78 = 2.34m

then, \frac{1}{2}kx² - mgx - mgh_o =0

 

\frac{1}{2}(530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

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3 years ago
What is the sprinters output at 2.0 s, 4.0 s and 6.0 s?
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A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?  

What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?  

Instantaneous Power is the force times velocity  

P = Fv  

Because the acceleration is constant, the force will be constant as well  

F = ma  

P = mav  

for constant acceleration, the velocity at each time is found using  

v = at  

P = ma(at) = ma²t  

find the acceleration using kinematic equation  

s = ½at²  

a = 2s/t²  

a = 2(47) / 7.0²  

a = 1.918 m/s²  

P(2.0) = 59(1.918²)2.0 = 434.25 W = 0.43 kW  

P(4.0) = 59(1.918²)4.0 = 868.51 W = 0.87 kW  

P(6.0) = 59(1.918²)6.0 = 1302.76 W = 1.3 kW  

I hope this helped.  


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How has globalization led to underdeveloped countries?
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Explanation: I hope this helps :)

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Explanation:

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