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Lyrx [107]
2 years ago
5

When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an

additional 0.90 cm. (a) what is the effective force constant of the springs? in N/m (b) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?
Physics
1 answer:
luda_lava [24]2 years ago
4 0

Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

x=100 cm=9*10^-3

effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

f=1/2*3.14√2.2*10^-5/310

f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

f=4.32*10^-5

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A rectangle has an area measuring 1350 square centimeters. Its length and width are whole numbers of centimeters. What are the p
Aleks04 [339]

The smallest perimeter of the rectangle is of value 150 cm.

Given:

The area of the rectangle, A = 1350 cm²

Calculation:

Let the length of the rectangle be 'x'

      the breadth of the rectangle be 'y'

We know that the area of a rectangle is given as:

A = (x) × (y)  

Applying values in the above equation we get:

xy = 1350 cm²

Factorizing the value of 1350, the possible values of length and breadth of the rectangle is as listed below:

x (cm)         y (cm)

1350      ×       1

675       ×       2

450       ×       3

270       ×       5

225       ×       6

150        ×       9

135        ×       10

90         ×       15

75         ×        18

54         ×        25

50         ×        27

45         ×        30         (least possible value)

Thus, the smallest perimeter of the rectangle can be calculated as:

P = 2 (x + y)

  = 2 (45 + 30)

  = 150 cm

Therefore, the smallest perimeter that the rectangle will have is 150 cm.

Learn more about factorization here:

<u>brainly.com/question/9231261</u>

#SPJ4

7 0
2 years ago
Helppppo!!!
goldfiish [28.3K]

Answer:

Electric force = 3.15\times 10^{-11}\textrm{ N}.

Explanation:

Given:

Charge on one grain, q_{1}=6\times 10^{-10}\textrm{ C}

Charge on second grain, q_{2}=2.3\times 10^{-15}\textrm{ C}

Distance between the two grains is, r=2\textrm{cm}=0.02\textrm{ m}

Electric force acting between two charges q_{1}\textrm{ and }q_{2} separated by a distance r is:

F_{e}=\frac{kq_{1}q_{2}}{r^2}

Where, k is Coulomb's constant equal to 9\times 10^9\textrm{ }Nm^2/C^2.

Now, plug in the given values and solve for F{e}.

F{e}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}

Therefore, the electric force between them is 3.15\times 10^{-11}\textrm{ N}.

5 0
3 years ago
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