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2 years ago

BRainliest if correct (if you don't know don't answer.) 2

Physics

2 answers:

klemol [59]2 years ago

6 0

Diagram B .... light shines through at an angle

max2010maxim [7]2 years ago

5 0

Diagram B

Explanation:

because when light passes from a denser to a rarer medium it bends towards the Principal axis due to refraction of light

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Use Kepler’s laws and the period of the Moon (27.4 d) to determine the period of an artificial satellite orbiting very near the

miv72 [106K]

As according to Kepler 's law

T =(4π²r³/ Gm)^1/2

here r= distance from earth center to satellite = 6400km = 6400000m

G = earth's gravitational constant= 6.67×10^-11

m = mass of earth= 5.98 ×10^24

so T =[ { 4× (3.1416)²×(6400000)³}/ {(6.67×10^-11)×( 5.98 ×10^24)} ]^1/2

T= 5133 sec

7 0

2 years ago

Read 2 more answers

To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially

cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside a pipe between two points.

Thus

⇒ $A_{1}$ × $V_{1}$ = $A_{2}$ × $V_{2}$

⇒ $\frac{\pi }{4}$ × $d_{1} ^{2}$ × $V_{1}$ =

⇒ $d_{1} ^{2}$ × $V_{1}$ = $d_{2} ^{2}$ × $V_{2}$

⇒ $(3.2)^{2}$ × 110 = $(0.57)^{2}$ × $V_{2}$

⇒ $V_{2}$ = 3467 cm / sec

⇒ $V_{2}$ = 34.67 m / sec

Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec

3 0

3 years ago

What’s the net force of an object being throwing into the air

timurjin [86]

Answer: forces acting on an object being thrown into the air is gravity and possibly air resistance

Explanation:

4 0

2 years ago

You are driving at the speed of 33.4 m/s (74.7296 mph) when suddenly the car in front of you (previously traveling at the same s

romanna [79]

Answer:

Part a)

a = - 8.45 m/s/s

Part b)

$d = 66 m$

Part c)

$d = 14.16 m$

Explanation:

Part a)

when car apply brakes then the friction force on the car in front of us is given as

$F_f = \mu mg$

here we need to find deceleration due to friction

$a = \frac{F_f}{m}$

$a = -\mu g$

$a = -8.45 m/s^2$

Part b)

Braking distance of the car is the distance that it move till it stops

so we will have

$v_f^2 - v_i^2 = 2ad$

$0 - 33.4^2 = 2(-8.45)d$

$d = 66 m$

Part c)

Since we know that average reaction time for human is 0.424 s

now we know that during reaction time our car will travel at uniform speed

so we will have

$d = vt$

$d = (33.4) (0.424)$

$d = 14.16 m$

4 0

2 years ago

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

2 years ago