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FrozenT [24]
3 years ago
14

Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe

ed into the standard
definition of power, P = W /delta t
Physics
2 answers:
Harman [31]3 years ago
8 0

Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -

                                       W=[ Fcos\theta]*S  where    Fcos\theta is the component of the force along the direction of displacement.

                                  Hence\ W= FScos\theta

                                                        = F.S

As per the question the power P is given as -

                                                  P=\frac{W}{\delta t}

                                                         =\frac{F.S}{\delta t}

                                                         = F.\frac{S}{\delta t}

                                                         = \ F.V

Hence alternative definition of power P = F.V


zimovet [89]3 years ago
5 0
Power formula is P= W/ Delta t, or we know that W= saclar product of D and F, D=length of distance traveled, F= the applied force, it is W = //D// //F// cos(D, F), when we substitute, we get P=[ //D// //F// cos (D, F)]/ deltat, or //D// = V x delta t, P= [V x delta t x //F// cos(D, F)] / delta t, and then P= V//F//cos(D, F), or V and D have the same direction, so cos(D, F)= cos(V, F), finally  P= V. F (or scalar product of V and F)
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A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
3 years ago
what is the acceleration produced by the resultant force acting on an object if the coefficient of friction acting between the b
LenKa [72]

The acceleration of the block is -0.98 m/s^2

Explanation:

The expression for the force of friction acting on the block is (assuming the surface is horizontal and flat):

F_f = -\mu mg

where

\mu is the coefficient of friction

m is the mass of the block

g is the acceleration of gravity

and where the negative sign means the direction of the force is opposite to that of the motion of the block

If this is the only force acting on the object, then this is also the resultant force, so we can rewrite Newton's second law as:

F=ma\\-\mu mg = ma

where

a is the acceleration of the block

Re-arranging the equation,

a=-\mu g

And so this is the expression for the acceleration of the block acted upon the force of friction.

In this problem, we have:

\mu=0.1

g=9.8 m/s^2

Solving,

a=-(0.1)(9.8)=-0.98 m/s^2

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

5 0
3 years ago
A soccer ball rolls to a stop of 21 m/s with a deceleration of -2.0 m/s. How far does it roll until it stops?
Luba_88 [7]

Answer:110.25 m

Explanation:v^2 = u^2 + 2as

21^2 = 0 - 2*2*s

s = 441/4

s=110.25 m

7 0
3 years ago
Read 2 more answers
The record height of a man to date is 8 feet 11 inches​ (107 inches). If all men had identical body​ types, their weights would
Alisiya [41]

Answer:

679.08 lbs

Explanation:

Here it is given that

m\propto h^3

m_1 = Mass of first person = 175 pounds

h_1 = Height of first person = 70 inches

m_2 = Mass of second person

h_2 = Height of second person = 110 inches

\frac{m_1}{m_2}=\frac{h_1^3}{h_2^3}\\\Rightarrow m_2=\frac{m_1h_2^3}{h_1^3}\\\Rightarrow m_2=\frac{175\times 110^3}{70^3}\\\Rightarrow m_2=679.08\ lb

The weight of the second person would be 679.08 lbs

5 0
3 years ago
A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2
Rudiy27

Answer:

Approximately 7.66\; \rm m.

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of u = 1.75\; \rm m \cdot s^{-1} and came to a stop when her speed became v = 0\; \rm m \cdot s^{-1}. How much time would that take if her acceleration is a = -0.20\; \rm m \cdot s^{-1}?

\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}.

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m.

In other words, the skateboarder travelled 15.3\; \rm m up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a},

where:

  • u and v are the initial and final velocity of the object,
  • a is the constant acceleration that changed the velocity of this object from u to v, and
  • x is the distance that this object travelled while its velocity changed from u to v.

For the skateboarder in this question:

\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}.

6 0
4 years ago
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