Both physical feature and behaviors can be inherited True False

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

rosijanka [135]

Answer:

A) $\frac{g}{b}(1-e^{-bt})$

Explanation:

Since a = g - bv,

We can substitute a = dv/dt into the equation.

Then, the equation will be like dv/dt = g - bv.

So we got first order differential equation.

As known, v = 0 at t = 0, and v = g/b at t = ∞.

Since $\frac{dv}{dt}= g - bv = b( \frac{g}{b} - v)$ ⇒ $\frac{dv}{ \frac{g}{b} - v}= bdt$

So take the integral of both side.

$- ln (\frac{g}{b} - v) = bt + C$

Since for t=0, v = 0 ⇒ $C =- ln (\frac{g}{b})$

$v = \frac{g}{b} + e^{-bt-ln(\frac{g}{b})} = \frac{g}{b}- \frac{g}{b}e^{-bt} = \frac{g}{b}(1-e^{-bt})$

5 0

3 years ago

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

Mrrafil [7]

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation $\theta=3.0\times10^{-5}\ rad$

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

$\sin\theta=\dfrac{1.22\lambda}{d}$

$\lambda=\dfrac{d\sin\theta}{1.22}$

Put the value into the formula

$\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}$

For small angle $\sin\theta\approx\theta$

$\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}$

$\lambda=4.92\times10^{-7}\ m$

$\lambda=492\ nm$

Hence, The maximum wavelength is 492 nm.

5 0

3 years ago

A uniformly charged conducting sphere of 1.3 m diameter has a surface charge density of 8.1 µc/m2. (a) find the net charge on th

8_murik_8 [283]

The surface charge density = q/A So q = surface charge density x Area The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998. So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6) The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space. So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)

8 0

3 years ago

You have been asked to make a roller coaster more exciting. The owners want the speed at the bottom of the first hill doubled. H

qaws [65]

Answer:

The height will be 4 times.

Explanation:

Given that,

The speed at the bottom of the hill doubled.

We need to calculate the height

Using conservation of energy

$K.E_{t}+P.E_{t}=K.E_{b}+P.E_{b}$

$K.E_{b}=P.E_{t}$

$\dfrac{1}{2}mv^2=mgh$

$\dfrac{1}{2}m(4v^2)=mgh$

Therefore,

$P.E = mg(4h)$

Here, m and g are constant

Hence, The height will be 4 times.

4 0

3 years ago

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi

Lelu [443]

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

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This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

4 0

1 year ago