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Ksenya-84 [330]
3 years ago
9

If an equation is dimensionally correct is thar equation a right equation​

Physics
1 answer:
Nady [450]3 years ago
8 0

If an equation is dimensionally correct, it does not mean that the equation must be true. On the other hand, when the equation is dimensionally correct, the equation cannot be true. Dimensional analysis is a technique used to check whether a relationship is correct

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Determine the ratio of the resistivity of pure water to silver?
shutvik [7]
Silver is a very good conductor, this means its resistivity is very low (from table, we can check the precise value, which is \rho_s = 1.6 \cdot 10^{-8} \Omega m).

Pure water, instead, is a very bad conductor, this means its resistivity is very high, of order of k \Omega \cdot m (10^3 \Omega m ). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
r= \frac{10^3 \Omega m}{10^{-8} \Omega m}  \sim 10^{11}

Therefore, we can say that the correct answer is
3 \cdot 10^{11} : 1
3 0
3 years ago
Look at the figure, which shows the motion of three balls. The curved paths followed by balls B and C are examples of _____. pro
Natali [406]

Answer:did you get the answer?

Explanation:

3 0
4 years ago
Read 2 more answers
Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp
Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*+\frac{1}{2} *2*8^{2}

x_{f}=-48+64\\x_{f}=16

5 0
4 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli
Setler79 [48]

Answer:

0.72

Explanation:

T = Time period of oscillation = 1.5 s

Angular frequency is given as

w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s

A = Amplitude of oscillation = 40 cm = 0.40 m

\mu = Coefficient of static friction = ?

a = acceleration of the block

m = mass of the block

Maximum acceleration of the block is given as

a = Aw^{2}

frictional force is given as

f = \mu mg

As per newton's second law

f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72

8 0
3 years ago
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