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sukhopar [10]
3 years ago
6

How much work is done when a very large force is applied to an object but does not move the object?

Physics
2 answers:
yarga [219]3 years ago
7 0

Answer:

A. None

(took the test)

Ierofanga [76]3 years ago
4 0
                         Work  =  (force) x (distance)

If the distance is zero (the object doesn't move), then it
doesn't matter how great the force is.  The work is still zero.
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Find the area of the part of the plane 5x 3y z = 15 that lies in the first octant.
Lunna [17]

This part of the plane lies above a triangle with boundaries x=0 and y=0 along the coordinate axes, as well as the line

z=0 \implies 5x + 3y = 15 \implies y = \dfrac{15 - 5x}3

When y=0, we have 15-5x=0\implies x=3. So this triangle is the set

T = \left\{(x,y)\in\Bbb R^2 ~:~ 0\le x\le3 \text{ and } 0\le y\le\dfrac{15-5x}3\right\}

Also, when x=0, we have y=\frac{15}3=5. So the triangle has length 3 and width 5, hence area 1/2•3•5 = 15/2.

Let z=f(x,y) = 15 - 5x - 3y. Then the area of the plane over T is

\displaystyle \iint_T dA = \iint_T \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy

We have

\dfrac{\partial f}{\partial x} = -5 \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 25

\dfrac{\partial f}{\partial y} = -3 \implies \left(\dfrac{\partial f}{\partial y}\right)^2 = 9

\implies\displaystyle \iint_T dA = \sqrt{35} \iint_T dx\,dy = \boxed{\frac{15\sqrt{35}}2}

since the integral

\displaystyle \iint_TdA

is exactly the area of T, 15/2.

8 0
2 years ago
If a car goes at a speed of 80 kph in 10 seconds, what is the distance
Blababa [14]

Answer:222 meters

Explanation:

Speed=80kph=(80x5/18)=(80x5)/18=22.2m/s

Time=10 seconds

distance=speed x time

distance=22.2 x 10

distance=222

distance=222 meters

3 0
4 years ago
a crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one
prohojiy [21]

The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be 37^oC is 367.42 Hz.

A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.

The fundamental frequency in the tube is given by

f=\frac{v_T}{4L}

where, v_T=v\sqrt{\frac{T}{273} }

Since, T=37+273 K = 310 K

v = 331 m/s

\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s

Using this, we get:

f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz

Hence, the fundamental frequency is 367.42 Hz.

To learn more about Attention here:

brainly.com/question/14673613

#SPJ4

7 0
2 years ago
This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me
Alexxx [7]

a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

Frequency = 9/5 = 1.8 oscillations per second

Frequency = 1.8 Hz

Period = 1/frequency = 1/1.8

Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

c) The formula for calculating the period of an ideal pendulum anywhere is

T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

Thus,

acceleration due to gravity = 25.56 m/s^2

Recall, earth's gravity = 9.8 m/s^2

number of g forces = 25.56/9.8

number of g forces = 2.61

6 0
1 year ago
When celery is placed in a glass of pure water the solution inside its cells is _____ compared to the water.?
kow [346]
The correct answer to the question above is hypertonic. When celery is being placed in a glass of pure water, the solution inside its cells is going to appear as hypertonic compared to the water. This means that the cells inside has a higher concentration than outside.
4 0
4 years ago
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