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2 years ago

A 7.0 kg box is at rest on a table. The static friction coefficient jis between the box and table is 0.40, and

Physics

1 answer:

Neko [114]2 years ago

8 0

Answer:

A m 23 3.3 32 m 0.37

Explanation:

So ur still thinking of me Just like I know you should. I cant not give you everything You know I wish I could.

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Which galaxy is the most stretched out?<br>

RideAnS [48]

Answer:

this is the anwser

Explanation:

The oddball spiral galaxy, called Messier 66, is one-thirdof the Leo Triplet, a group of three interacting galaxies about 35 millionlight-years from Earth (a light-year is the distance light can cover in ayear).

4 0

3 years ago

Read 2 more answers

While watching a planet orbiting a star, you notice that the doppler shift in the light of the star is becoming increasingly blu

Lyrx [107]

Answer:

The star must be moving "closer" to the earth because the wave fronts that are being emitted are shifted toward the blue end of the spectrum (they are closer together)

6 0

2 years ago

Bicycle in its writers have combined a mass of 80 kg in a speed of 6.0 meters per Second what is the magnitude of the average fo

erik [133]

Answer:

F = 120 N

Explanation:

Force x distance = energy

The bike has energy 1/2 . 80 . 6^2 = 1440 J

You are looking at an example of not reading the question properly.

Impulse = Force . time = change in momentum

F . 4 = 80 .6

F = 120 N

7 0

3 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

3 years ago

Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po

coldgirl [10]

Answer:

the intensity of the light after passing through the two polarizing filters is 4.11 units

Explanation:

Given the data in the question;

the intensity of an unpolarized light; I₀ = 25.0 units

when the unpolarized light passes through the first polarizer, its intensity reduces to half of its initial value;

⇒ I₁ = I₀/2 = 25/2 = 12.5 units

the angle between the transmission axes of two polarizers is;

∅ = 55° - 0° = 55°

The intensity of the light after passing through two polarizing filters will be;

I₂ = I₁cos²∅

we substitute

I₂ = 12.5 × cos²(55)

I₂ = 12.5 × 0.3289899

I₂ = 4.11 units

Therefore, the intensity of the light after passing through the two polarizing filters is 4.11 units

6 0

2 years ago