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Anuta_ua [19.1K]
3 years ago
11

You ordered a large block of wood with length 5, width 2, and height 1 (each in feet). Unfortunately, the manufacturer can only

guarantee to meet these measurements up to an error of 0.1 excess feet per side. Use a suitable gradient to approximate the most total excess volume that may occur. Compute the exact maximal excess volume.
Physics
1 answer:
Gennadij [26K]3 years ago
5 0
If each side is 0.1 feet extra,
The volume will be 5.1*2.1*1.1= about 11.781.
Perhaps this helps.
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A ray diagram shows that an object is placed in front of a plane mirror. What are the characteristics of the image produced by t
SVEN [57.7K]

D) upright , same size as object ,virtual

3 0
3 years ago
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A machine is supplied Energy at a rate of 4000 W and does useful work at a rate of 3760 W. What is the efficiency of the machine
ICE Princess25 [194]
The efficiency of a machine is the ratio between the output power and the power in input:
\epsilon =  \frac{P_{out}}{P_{in}}
in our problem, the output power is 3760 W while the input power is 4000 W, so the efficiency is
\epsilon =  \frac{3760 W}{4000 W} =0.94
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6 0
3 years ago
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I NEED HELP ASAP!!!!!
Ratling [72]

Answer:

D) momentum of cannon + momentum of projectile= 0

Explanation:

The law of conservation of momentum states that the total momentum of an isolated system is constant.

In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:

p_i = p_f

But the initial momentum is zero, because at the beginning both the cannon and the projectile are at rest:

p_i = 0

So the final momentum, which is sum of the momentum of the cannon and of the projectile, must also be zero:

p_f = p_{cannon}+p_{projectile} =0

6 0
4 years ago
I'm very confused. Thanks for whoever helps me :)
sergij07 [2.7K]
(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**
6 0
3 years ago
How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius
Korvikt [17]

Answer:

<h2>The coefficient of static friction will be 0.7</h2>

Explanation:

Given data

the radius of curve= 90m

speed v= 90 km/h to m/s = (90*100)/60*60=  25 m/s

we know that the expression for the centripetal force acting on the car

Fc= \frac{mv^2}{r}-------1

we also know that the expression for the frictional force between road and tire.

Ff= μmg--------2

Equating equation 1 and 2 we have

μmg= mv^2/r

μ= v^2/gr

substituting the values of speed and radius we have (assuming g= 9.81m/s^2)

μ= 25^2/9.81*90

μ= 625/882.9

μ= 0.7

4 0
3 years ago
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