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Kaylis [27]
3 years ago
14

Which of the following is an expression of Newton's second law? A. The acceleration of an object is determined by its mass and t

he net force acting on it. O O B. Objects at rest tend to stay at rest and objects in motion tend to stay in motion unless acted on by an unbalanced force. C. Objects at rest tend to stay at rest, and objects in motion tend to slow down until they come to rest. D. Every force is paired with an equal and opposite force.​
Physics
1 answer:
Sav [38]3 years ago
7 0

Answer: A. The acceleration of an object is determined by its mass and the net force acting on it.

Explanation:

Newton's second law of motion explains that the acceleration of an object will depend on two vital variables which are the mass of the object and the net force that's acting on it.

It should be noted that the acceleration of the object directly depend on the net force while it depends inversely on the mass. Therefore, when the force that's acting on such object is increased, then the acceleration will increase as well. On the other hand, when there is an increase in mass, there'll be a reduction in the acceleration.

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A balloon of hydrogen is put into to pressure chamber. The initial pressure and volume of hydrogen is 1 atm and .5 cm3. The pres
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Answer:

0.25 cm³.

Explanation:

We shall apply Boyle's law to find the solution . According to it

PV = constant where P is pressure and V is volume of the gas.

P₁ V₁ = P₂V₂

1 x .5 = 2 x V₂

V₂ = 0.25 cm³.

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3 years ago
Calculate the mass of an object that has a momentum of 100kg x m/sec and velocity of 4 m/sec
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P= mv

where p is momentum
m is mass
v is velocity

so it's given p= 100kgm/sec
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so putting in the formula

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A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in rad
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Answer:

\displaystyle \theta' =0.24\ rad/s

Explanation:

<u>Rate Of Change</u>

Let some variable y depend on time t. we can express y as a function of t as

y=f(t)

The instant rate of change of y respect to t is the first derivative, i.e.

y'=f'(t)

The balloon, the ground and the observer form a right triangle (shown below) where the height of the balloon y, the horizontal distance x, and the angle of elevation are related with the trigonometric formula

\displaystyle tan\theta =\frac{y}{x}

Since x is constant, we take the derivative with respect to time  by using the chain rule:

\displaystyle sec^2\theta \ \theta' =\frac{y'}{x}

Solving for \theta'

\displaystyle \theta' =\frac{y'}{xsec^2\theta}

Let's compute the actual angle with the initial conditions y=9 feet, x=12 feet

\displaystyle tan\theta =\frac{y}{x}

\displaystyle tan\theta =\frac{9}{12}=\frac{3}{4}

Knowing that

\sec^2\theta=1+tan^2\theta

\displaystyle \sec^2\theta=1+\left(\frac{3}{4}\right)^2

\displaystyle \sec^2\theta=\frac{25}{16}

The balloon is rising at y'=8 feet/sec, thus we compute the change of the angle of elevation:

\displaystyle \theta' =\frac{8}{12\ \frac{25}{16}}

\displaystyle \theta' =\frac{32}{75}\ rad/s

\boxed{\displaystyle \theta' =0.43\ rad/s}

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