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3 years ago

13

fins the powee of a convex lens which forms a real and inverted image of magnification -1 of an object placed at a distance from

its optical centre

1 answer:

anzhelika [568]3 years ago

7 0

Explanation:

hope i am correct

hii

how are you

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I need help with question 4

stellarik [79]

See if it’s answer A

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3 years ago

Plz help me!!!! A spring is connected to a wall as shown below. A mass on a horizontal surface is connected to the springs and p

kirill [66]

Answer:

320 N/m

Explanation:

F = k·Δx

where

F is the restoring force of the spring

k is the proportionality constant called the ‘spring constant’

Δx is the change in the spring’s position due to the deformation.

You need the k so

25 cm= 0.25m

k=F/∆x = 80/0.25 = 320 N/m

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3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

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3 years ago

If the mass of a material is 50 grams and the volume of the material is 5 cm^3, what would the density of the material be?

kotykmax [81]

D = M / V

D = 50 / 5

D = 10 g/cm^3

5 0

3 years ago

Read 2 more answers

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

alisha [4.7K]

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

7 0

3 years ago

Read 2 more answers