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kiruha [24]
2 years ago
10

A 5 inch tall balloon shoot doubles in height every 3 days. if the equation y=ab^x, where is x is the number of doubling periods

,represents the height of the bamboo shoot,what are the values of a and b?
Physics
1 answer:
VashaNatasha [74]2 years ago
3 0
In this item if we let x be the number of doubling period, that is number of days divided by 3, and y be the height of the bamboo shoot, in the equation
                                     y  = ab^x
The value of a is 5 because that is the original height of the shoot and b is 2 because we are talking about the doubling rate. The equation is therefore,
                                    y = 5(2^x)
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Explain why ice cubes formed from water of a glacier freeze at a higher temperature than ice cubes
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Because icebergs contains more molecules, and tempature measures. Heat is the energy of that motion. So there forward because there are more molecules in a iceberg then a ice cube
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2 years ago
Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment
riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a  look at the free bodily diagrammatic representation.

The first diagram will aid us in answering  question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ  = \frac{360+240}{450}

tan θ  = \frac{600}{450}

tan θ  = 1.333

θ  = tan⁻¹ (1.333)

θ  = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

M__B}=(Fcos \theta *240)-(Fsin \theta *450)

where Force(F) = Force in the cord AC = 1350 N and θ  = 53.13° ; we have:

M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)

M__B}= 194400.463-485999.348

M__B}=-291598.885 N-mm\\

M__B}=-291.598 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B (M__B}),

we have:

M__B}=-Fcos \theta *360 - Fsin \theta * 0

M__B}=-Fcos \theta *360 - 0

M__B}=-Fcos \theta *360

where Force(F) =  1350 N and θ  = 53.13° ; we have:

M__B}= -1350*cos53.13^0*360

M__B}= -291600 N-mm

M__B}= -291.6 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

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2 years ago
A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow
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Answer:

Explanation:

Given

Force P is acting upward

C is vertical contact Force

W is the weight of the crate

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P-W+C=0

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7 0
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A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

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k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

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q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C

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brainly.com/question/8960054

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20.3 divided by 3.0 will get u velocity and v times 3.0s 
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