The rate of doing work is called _____.
power
force
energy
inertia

Answers

Answer 1

Answer:

The rate of doing work is called power.

Hence, the correct option is A.

Power is a measure of how quickly or how much work is done per unit of time.

It is the amount of energy transferred or converted per unit time. The unit of power is the watt (W), where 1 watt is equal to 1 joule of work done per second.

Therefore, The rate of doing work is called power.

Hence, the correct option is A.

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Answer 2

Answer:

Final answer:

Power is the correct term in physics for the rate of doing work. It signifies the amount of work done or energy converted per unit of time. Other options provided like force, energy, and inertia have different meanings in physics.

Explanation:

The rate of doing work is called power. In physics, power is defined as the amount of work done or energy converted per unit of time. For example, when you lift a heavy box, you're doing work, and the faster you lift that box (maintaining the same force), the more power you're exerting.

Whereas force is a push or pull upon an object resulting from the object's interaction with another object, energy is the capacity for doing work, and inertia refers to an object's resistance to change in motion.

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State the relationship between degrees Celsius and kelvins

Answers

Kelvin is a unit of absolute temperature and is often used in measurements and it can also be associated with degrees Celsius. 0 Kelvin is equivalent to -273.15 degrees Celsius.

In order to solve for either unit of temperature, the equation is:

T(Kelvin)= T(degrees Celsius) + 273.15
T(degrees Celsius) = T(Kelvin) - 273.15

A car accelerates uniformly from rest to a speed of 58.3 mi/h in 9.47 s. Find the constant acceleration of the car.

Answers

58.3 mi/h = (58.3x1605)/3600= 24 m/s
Acceleration = 24/9.47= 2.5 m/s^2

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the acceleration a of the object is given by a = g bv, where v is the object's speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time? A) v = g(1-e^-bt)/b B) v = (ge^bt)/b C) v = (g+a)t/b

Answers

Answer:

A) $(g)/(b)(1-e^(-bt))$

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 $(dv)/(dt)= g - bv = b( (g)/(b) - v)$ ⇒ $(dv)/( (g)/(b) - v)= bdt$

So take the integral of both side.

$- ln ((g)/(b) - v) = bt + C$

Since for t=0, v = 0 ⇒ $C =- ln ((g)/(b))$

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

The correct option for the expression of speed as an explicit function of time is option A

A) v = g·(1 - $e^{-b \cdot t$ )/b

The reason why option A is correct is given as follows;

Known:

The initial velocity of the object at time t = 0 is v = 0 (object at rest)

The function that represents the acceleration is a = g - b·v

Where;

v = The speed of the object at the given instant

b = A constant term

By considering the limiting case for time t, we have;

At very large values of t, the velocity will increase such that we have;

$\lim \limits_(t \to \infty) a = 0$ therefore, $\lim \limits_(t \to \infty) g - b\cdot v = 0$ and $\lim \limits_(t \to \infty) \left( v_(max) = (g)/(b) \right)$

The given equation can be rewritten as follows, to express the equation in terms the velocity;

$a = b \cdot \left((g)/(b) - v \right) = b \cdot \left(v_(max) - v \right)$

$Acceleration, \ a = (dv)/(dt)$

Therefore;

$(dv)/(dt) = b \cdot \left((g)/(b) - v \right)$

The above differential equation gives;

$(dv)/( \left((g)/(b) - v \right)) = b \cdot dt$

Which gives;

$\displaystyle \int\limits {(dv)/( \left((g)/(b) - v \right)) } = \int\limits {b \cdot dt} = b \cdot t + C$

$\displaystyle \int\limits {(dv)/( \left((g)/(b) - v \right)) } = -\ln \left((g)/(b) - v \right)$ and $\displaystyle\int\limits{b \cdot dt} = b \cdot t + C$

Therefore

$\displaystyle -\ln \left((g)/(b) - v \right) =b \cdot t + C$

At t = 0, v = 0, therefore;

$\displaystyle -\ln \left((g)/(b) - 0 \right) =b * 0 + C$

$C = \displaystyle -\ln \left((g)/(b) \right)$

Which gives;

$\displaystyle -\ln \left((g)/(b) - v \right) =b \cdot t \displaystyle -\ln \left((g)/(b) \right)$

$\displaystyle \ln \left((g)/(b) - v \right) =-b \cdot t \displaystyle +\ln \left((g)/(b) \right)$

$\displaystyle (g)/(b) - v = e^{-b \cdot t \displaystyle +\ln \left((g)/(b) \right)} = e^(-b \cdot t) * e^\ln \left((g)/(b) \right)} = e^(-b \cdot t) * (g)/(b)$

$\displaystyle (g)/(b) - e^(-b \cdot t) * (g)/(b) = v$

$\displaystyle (g)/(b) \cdot \left(1 - e^(-b \cdot t) \right) = v$

∴ v = g·(1 - $e^{-b \cdot t$ )/b

The correct option is option (A)

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A 5 kg box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force of 17 newtons west. What is the magnitude of the acceleration of the box? Type only numbers. Do not include the units of m/s2.

Answers

Answer:

The acceleration of the box is 2.

Explanation:

According to Newton's second law of motion, the acceleration of any object will be directly proportional to the net unbalanced force acting on the object and inversely proportional to the mass of the object.

Net force = Mass × Acceleration

So $Acceleration = (Net force)/(Mass)$

Since in this case, the box is experiencing a force from east of magnitude 27 N and resisting force of about 17 N from west. So the net force will be the difference of acting and reacting force.

Net force = 27-17 = 10 N.

Thus, $Acceleration = (10 N)/(5 kg)$

So 2 $m/s^(2)$ is the acceleration of the box. Thus the magnitude of acceleration of the box is 2.

What is a scientific paradigm

Answers

a scientific paradigm is is a distinct set of concepts or thought patterns, including theories, research methods, postulates, and standards for what constitutes legitimate contributions to a field.

List all physical changes.

Answers

Physical changes refer to alterations in the properties or characteristics of matter without changing its chemical composition. Some examples of physical changes include changes in state, size, shape and color.

Changes in state are common physical changes that occur when matter transitions between solid, liquid, or gas phases. This change occurs due to a change intemperature or pressure. For example, water freezing into ice or boiling into steam are examples of changes in state.

Changes in size, shape, and texture occur when matter undergoes physical alterations without any changes in its chemical composition. For instance, cutting a piece of paper into small pieces, hammering a metal sheet to make it thinner, or molding clay into a different shape are all examples of physical changes.

Changes in color are also physical changes. They occur when the light-reflecting properties of a material change due to a physical alteration. For instance, when copper is exposed to air, it turns green due to a chemical reaction, but when iron rusts, it changes color due to a physical change in its surface.

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