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stealth61 [152]
3 years ago
6

A concrete block is hung from an ideal spring that has a force constant of 100 N/m . The spring stretches 0.129 m .A- What is th

e mass of the block? B- What is the period of oscialltion of the block if it is pulled down 1 cm and released? C- What would be the period of oscillation if the block and spring were placed on the moon?
Physics
1 answer:
madam [21]3 years ago
8 0

Answer:

1.31498 kg

0.72050 s

0.72050 s

Explanation:

m = Mass of block

g = Acceleration due to gravity = 9.81 m/s²

k = Spring constant = 100 N/M

x = Displacement = 0.129 m

The force balance is

mg=kx\\\Rightarrow m=\dfrac{kx}{g}\\\Rightarrow m=\dfrac{100\times 0.129}{9.81}\\\Rightarrow m=1.31498\ kg

The mass of the block is 1.31498 kg

Time period is given by

T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{1.31498}{100}}\\\Rightarrow T=0.72050\ s

The period of oscillations is 0.72050 s

The time period does not depend on the acceleration due to gravity. It varies with the mass and the spring constant.

Hence, the time period would be the same

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F(normal force) = MA * cos(angle)

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Don't have a calculator, so can't really do all the math right there. So just plug that in

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7 0
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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
What is a relatively soft layer of upper mantle called?
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3 0
3 years ago
Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
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' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

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= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0
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