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4 years ago

14

It is known that a certain kind of algae in the Dead Sea can double in population every 2 days. Suppose that the population of a

lgae grows exponentially, beginning now with a population of 1,000,000. (a) How long it will take for the population to quadruple in size?

1 answer:

Hoochie [10]4 years ago

8 0

Answer:

4 days

Explanation:

1,000,000*4=4,000,000(total size after quadrupling)

1,000,000*2=2,000,000(after 2 days.)

2,000,000*2=2,000,000(after 4 days.)

therefore it will take 4 days to quadruple in size

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Escribo dentro del paréntesis la letra (F) si el enunciado es falso o (V) si el enunciado es verdadero. Fundamento si mi elecció

KatRina [158]

Answer:

El primero es (V) y el segundo es(f)

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3 years ago

Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used

natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

At 600 MPa of stress

The strain is $0.3mm/mm$

At 450 MPa of stress

The strain is $0.0015 mm/mm$

To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

$=1398.6N$

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

$T = 6 * F$

$= 6* 1398.6$

$T =8391.6 N$

6 0

4 years ago

Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At w

Alisiya [41]

Answer:

C. At the instant the ball reaches its highest point.

Explanation:

When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"

Remember, F = ma = m(v/t)

Since v = 0 at maximum height

F = m(0/t)

F = 0N

This shows that the force acting on the body is zero at the maximum height.

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3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

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so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

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3 years ago

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Mariulka [41]

Answer:

Acceleration

Explanation:

Its speed or velocity change

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3 years ago