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CaHeK987 [17]
3 years ago
8

When you blow up a balloon, what happens to the volume of air inside the balloon ?

Physics
1 answer:
shutvik [7]3 years ago
3 0
<span>If the balloon is closed then both volume and pressure will increase when the gas inside is heated.</span>
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Derive the kinetic equations for Vmax and KM using the transit time and net rate constant method as discussed in class and follo
Pepsi [2]

Answer:

Rate = vmax k3/k2+k3

Explanation:

The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.

This is usually expressed as the Km ie. Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.

Please kindly check attachment for the step by step solution of the given problem.

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3 years ago
Which of the following is not a concern regarding waste
juin [17]
I believe it is noise pollution

4 0
3 years ago
A spherical shell has inner radius 1.5 m, outer radius 2.5 m, and mass 850 kg, distributed uniformly throughout the shell. What
Nimfa-mama [501]

Answer:

The magnitude of the gravitational force is 4.53 * 10 ^-7 N

Explanation:

Given that the magnitude of the gravitational force is F = GMm/r²

mass M = 850 kg

mass m = 2.0 kg

distance d = 1.0 m , r = 0.5 m

F = GMm/r²

Gravitational Constant G = 6.67 × 10^-11 Newtons kg-2 m2.

F = (6.67 × 10^-11 * 850 * 2)/0.5²

F = 0.00000045356 N

F = 4.53 * 10 ^-7 N

3 0
3 years ago
In a given circuit, the supplied voltage is 1.5 volts. One resistor is 3 ohms and the other is 2 ohms. Use Ohm's law to determin
tino4ka555 [31]
I = E / R

If the resistors are in series, the current is 0.3 Amperes.

If the resistors are in parallel, the current is 1.25 Amperes.
4 0
3 years ago
Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

8 0
3 years ago
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