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andreev551 [17]
3 years ago
13

What is the dimensional formula of young modulas​

Physics
2 answers:
LekaFEV [45]3 years ago
6 0

Answer:

The dimensional formula of Young's modulus is [ML^-1T^-2]

natita [175]3 years ago
4 0

Answer:

G.oogle : The dimensional formula for Young’s modulus is:

A. [ML−1T−2]A. [ML−1T−2]

B. [M0LT−2]B. [M0LT−2]

C. [MLT−2]C. [MLT−2]

D. [ML2T−2]

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Pls help i’m in a test and don’t know what to do
Tresset [83]

Answer:

I think its C, even with the typo.

6 0
3 years ago
Obtain the formula for the focal length of a lens in terms of object distance (u)
Alexxx [7]

Answer:

m=image distance÷object distance

6 0
2 years ago
P= F/A make a the subject
lawyer [7]

Answer:

a=f/p

Explanation:

p=f/a

f=p×a

this all

4 0
3 years ago
The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively.(a) Calculate the force of attraction between these tw
timurjin [86]

Answer:

1.09527\times 10^{-8}\ N

Explanation:

q_1 = Mg ion = +2q

q_2 = F ion = -q

q = Charge of electron = 1.6\times 10^{-19}\ C

r = Distance between ions = 0.072+0.133\ nm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electrical force is given by

F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N

The attractive force is 1.09527\times 10^{-8}\ N

8 0
4 years ago
A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as
Leona [35]

(a) -48.0 cm, diverging

We can use the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the object distance

q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)

Solving for f, we find the focal length of the lens:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{-12.0 cm}=-0.021 cm^{-1}

f=\frac{1}{-0.021 cm^{-1}}=-48.0 cm

The lens is diverging, since the focal length is negative.

(b) 6.38 mm, erect

We can use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the size of the image

y = 8.50 mm is the size of the object

Substituting p and q that we used in the previous part of the problem, we find y':

y'=-y\frac{q}{p}=-(8.50 mm)\frac{-12.0 cm}{16.0 cm}=6.38 mm

and the image is erect, since the sign is positive.

(c)

See attached picture.

4 0
3 years ago
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