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Greeley [361]
3 years ago
15

Two springs with spring constants k1 and k2 are connected together. With mass m1 attached, the springs have

Chemistry
1 answer:
Monica [59]3 years ago
7 0
K1x1=k2x2=F
k2x2/k1=x1
F=keqx=keq(x1+x2)

sub.s in for x1:F=keq(k2x2/k1+x2)

sub.s in for F (in terms of k2 and x2)k2x2=keq(k2x2/k1+x2)

x2 is in every term:k2=keq(k2/k1+1)

multiply both sides by (k1/k2+k1)
keq=(k1/k1+k2)k2
keq=(1/k1+1/k2)−1
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3 years ago
An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h
vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base  =  w

Height of the pyramid = h

Volume of the pyramid = V=\frac{1}{3}lwh

We have:

Rate at which water is filled in cube = \frac{dV}{dt}= 45 cm^3/s

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

V=\frac{1}{3}\times l^2\times h

\frac{l}{h}=\frac{6}{13}

l=\frac{6h}{13}

V=\frac{1}{3}\times (\frac{6h}{13})^2\times h

V=\frac{12}{169}h^3

Differentiating V with respect to dt:

\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}

\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}

Putting, h = 4 cm

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}

=13.20 cm/s

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0
3 years ago
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