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3 years ago

A sphere has______ sides

Physics

2 answers:

Alina [70]3 years ago

8 0

Answer:

5738929292

Explanation:

simple math

o-na [289]3 years ago

8 0

A sphere has no sides

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you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic

Tomtit [17]

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

$W=Fd$

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

$d=\frac{h}{sin \theta}$ is the distance covered, where

h = 1 m is the vertical rise

$\theta=30^{\circ}$ is the slope of the plane

Substituting and solving, we find

$W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J$

Learn more about work:

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#LearnwithBrainly

5 0

3 years ago

An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi

sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

now we know that

object distance from mirror is

$d_o = -50 cm$

Focal length of the mirror is given as

$f = 40 cm$

now we have

$\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}$

$\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}$

$d_i = 22.22 cm$

6 0

3 years ago

Which expression is equivalent to 5(2+7)? 0 2(5+7) O 2 + 7(5) O 5(2)+7 O 5(2)+5(7)

professor190 [17]

Answer:

5(2)+5(7)

Explanation:

5(2)+5(7)=10+35=45

6 0

3 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

<u>Explanation:</u>

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

4 0

4 years ago

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

marshall27 [118]

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

Let the linear charge density of the charged wire is given as

$\frac{q}{L} = \lambda$

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

$\int E. dA = \frac{q}{\epsilon_0}$

here we have

$E \int dA = \frac{\lambda L}{\epsilon_0}$

$E. 2\pi r L = \frac{\lambda L}{\epsilon_0}$

so we have

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

4 0

3 years ago

A sphere has______ sides​

A sphere has______ sides