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Anni [7]
3 years ago
10

What are the characteristics of the image based on the values?Check all that apply

Physics
2 answers:
Ahat [919]3 years ago
8 0

inverted, real, behind the lens is the answer

kifflom [539]3 years ago
7 0

Answer:

Inverted

Real

Behind the lens

Explanation:

As we know that

d_o = 16 cm

d_i = 16 cm

also we know that

h_o = 4 cm

h_i = -4 cm

now we know that magnification is given by

M = \frac{d_i}{d_o} = \frac{h_i}{h_o}

so we can say here that

M = 1

so here image size will be same as object size and it must be real and inverted as magnification is negative in sign

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A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas
Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the  block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface   and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block  : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the  block

F₁ = 65 N,  horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight  of the block

W= m*g  = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force  (FN)

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N=  0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax ,  ax= a  : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

4 0
3 years ago
A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud
Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride  6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

                                = 9.28 cos 27.27°

                                = 8.24 km

horizontal axis component = 9.28 sin ∅

                                             = 9.28 sin 27.27°

                                             = 4.5 km

distance of the final point from the origin

                            = 7.99 -(6.2-4.5)

                            = 6.29 km

displacement

d = \sqrt{6.29^2+8.24^2}

d = 10.37 km

b) tan \theta = \dfrac{6.29}{8.24}

θ = 37.36°

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ohaa [14]

Answer:

Cobalt

Explanation:

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