The force that pulls objects toward each other

Jenny pushes a 40 N crate down the hall 2m. How much work did she do?

Shalnov [3]

Work done = force x distance = 40 x 2 = 80 Joules.

8 0

3 years ago

a 5. A bar has a 20 N weight at one end. You have a weight of 15 N to hang somewhere on the bar so that the bar is in equilibriu

Dmitry_Shevchenko [17]

the answer is a

Explanation:

it's a because 5n time 4 will give u 20n

5 0

2 years ago

Derive a relation between critical angle and refractive index

Vera_Pavlovna [14]

Answer:

At the critical angle the angle of refraction (outside the medium) is 90°

Ni sin θ = Na sin 90° = 1 assuming outside medium is air

sin θ = 1 / Ni where Ni is the refractive of the medium involved

4 0

2 years ago

A cylindrical conductor with a circular cross section has a radius a and a resistivity p and carries a constant current I. (Take

Lady bird [3.3K]

Answer:

Explanation:

a)

Using Ohms Law

$R= \rho \frac{l}{\pi a^2 }
\\V = IR = E l = I \rho \frac{l}{\pi a^2 }
\\E = \frac{I \rho}{\pi a^2 }
\\E = J \rho
$

Where J is the current density $J = \frac{I}{\pi a^2 }
$

and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor $E = \rhoJ$ just inside at a radius a (and anywhere else).

b)

Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current

$\oint B .dl = B 2 \pi a = \mu_{o} I
$

such that

$B = \frac{\mu_{o} I}{2 \pi a }
$

and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.

c)

The Poynting vector is given by

$S = \frac{1}{ \mu_o} |E \textrm{x}B| = \frac{\rho I^2}{2 \pi^2 a^3}
$

and by the right hand rule it’s always pointing in towards the center of the conductor.

d)

Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c

7 0

3 years ago

A luggage handler pulls a 20.0 kgkg suitcase up a ramp inclined at 32.0 ∘∘ above the horizontal by a force F⃗ F→ of magnitude 16

Nuetrik [128]

Answer:

a) W₁ = 8242.2 J, b) W₁ = 8242.2 J , c) W₃ = 0 , d) W₄ = -189.51 J ,

f) v = 27.24 m / s

Explanation:

a) Work is defined by

W = F. d = F d sin θ

where angle is between force and displacement

n this case the suitcase is going up and the outside F is parallel to the plane, so the angle is zero and the cosine is 1

W = F d

Let's calculate

W = 169 3.8

W₁ = 8242.2 J

b) the gravitational force is vertical so it has an angle with respect to the horizontal parallel to the plane of

θ’= 90 - θ

θ'= 90-32 = 58º

W = m g d thing θ ’

W = 20 9.8 3.8 thing (180 + tea ’) =

W = 744.8 cos (180 + 32)

W₂ = -631.6 J

c) The normal work, as it has 90º with respect to the displacement, its work is zero

W₃ = 0

d) the work of the friction force

Let's write Newton's second law the Y axis

N- Wy = 0

Cos 32 = Wy / W

N = W cos 32

The expression for friction force is

fr = μ N

fr = μ mg cos 32

fr = 0.300 20 9.8 cos (32)

fr = 49.87 N

The work of the friction force

W = fr d cos 180

W₄ = -49.87 3.8

W₄ = -189.51 J

E) The total work

W = W₁ + W₂ + W₃ + W₄

W = 8242.2- 631.6 + 0 -189.51

W_total = 7421.09 J

F) Usmeosel theorem of work and energy

W = ΔK

W = ΔK = ½ m v² - 0

v =√ 2W / m

v = √ (2 7421.09 / 20)

v = 27.24 m / s

5 0

4 years ago