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3 years ago

Please answer this question sqdancefan

Physics

2 answers:

igor_vitrenko [27]3 years ago

5 0

Answer:

Explanation:

#CARRYONLEARNING

justoko2

AleksandrR [38]3 years ago

4 0

Answer:

(D) 4

Explanation:

The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.

density = mass/volume

density = mass/(π(radius^2)(length))

So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.

_____

If you look at the maximum and minimum density, you find they are ...

{0.0611718, 0.0662668} g/(mm²·cm)

The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.

The error is about 4%.

_____

<em>Additional comment</em>

If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.

Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.

You might be interested in

If the designers of submarines do not appropriately account for hydrostatic pressure,

andre [41]

Answer:

A) the pressure could crush the submarine

Explanation:

The submarines are mainly used for defense purposes.
In world war 1 and 2 the submarines played a vital role in the war.
The submarines work according to the principle of buoyancy.
According to this principle, when seawater is filled inside the tank in a required amount, the submarines sink.
When the seawater is expelled, the submarine rises to above the sea.
Most of the time, it dives deep into the sea, the hydrostatic pressure should be taken into account.
If the hydrostatic pressure is not considered, the pressure could crush the submarine.

7 0

3 years ago

What is the final velocity of a rocket that is traveling at 205 m/s and accelerates at

-Dominant- [34]

Answer:

70,100 m/s

Explanation:

Given:

v₀ = 205 m/s

a = 8.03 m/s²

Δx = 1750 m

Find: v

v² = v₀² + 2aΔx

v² = (205 m/s)² + 2 (8.03 m/s²) (1750 m)

v = 70,130 m/s

Rounding to three significant figures, the final velocity is 70,100 m/s.

7 0

4 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$