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Brut [27]
3 years ago
6

Which of the following classifications of star temperature is coolest?

Physics
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0
The coolest would be m the hottest would be o
Oliga [24]3 years ago
3 0
The hottest would be the O type and the coolest is M
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A) A car accelerates uniformly at 3.7 m s⁻² as it passes a stationary road train. The initial speed of the car is 30 m s⁻¹, and
Tanya [424]

(a) The length of the train is 54.6 m

(b)  The braking distance of the road train is 194.44 m

The given parameters:

acceleration of the car, a = 3.7 m/s²

initial velocity of the car, u = 30 m/s

final velocity of the car, v = 130 km/h = 36.11 m/s

To find:

  • the length of the train

The length of the train is the distance travelled by the car

The distance traveled by the car is calculated as:

v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\s = \frac{ v^2 - u^2}{2a} \\\\where;\\\\s \ is \ the \ distance \ travelled \ by \ the \ car\\\\s = \frac{(36.11)^2 - (30)^2}{2\times 3.7} \\\\s = 54.585 \ m\\\\s \approx 54.6 \ m

Thus, the length of the train is 54.6 m

(b) The braking distance of a road train travelling at 25 m s⁻¹

a = \frac{v_1^2}{2s_1} = \frac{v_2^2}{2s_2} \\\\Given;\\\\s_1 = 70 \ m\\\\v_1 = 15 \ m/s\\\\v_2 = 25 \ m/s\\\\s_2 = ?\\\\2s_2v_1^2 = 2s_1v_2^2\\\\s_2 = \frac{ 2s_1v_2^2}{2v_1^2} \\\\s_2 = \frac{s_1v_2^2}{v_1^2} \\\\s_2 = \frac{70\times 25^2}{15^2} \\\\s_2 = 194.44 \ m

Thus, the braking distance of the road train is 194.44 m

Learn more here: brainly.com/question/19572178

6 0
2 years ago
Objects in space are moving at a constant velocity in a straight line.
tigry1 [53]

Answer:

C

Explanation:

An object in motion will stay in motion unless acted on by a net positive or negative force.

For answer A. If the object were to be in an orbit, it would inevitably accelerate due to it being acted on by the gravitational force from the object it is orbiting. At different points in the orbit, the object will move at different speeds and continuously transfer between kinetic and potential energy.

For answer B. The object would would not stop their motion. In order for the object to lose energy, it would have to transfer it through friction or through its interaction with a gravitational field.

For answer D. No energy is "required" to maintain constant motion unless the object is willingly fighting against a resistive force like friction or a graviational well.

8 0
2 years ago
What is the science behind the making of pop-up books?
IRINA_888 [86]

Answer

The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.

Explanation

Hope that this helps you and have a great day:)

6 0
3 years ago
A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the wate
Vilka [71]

Answer:

   f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

        ∑ F = ma

where the acceleration is

         a = \frac{d^2 y}{dt^2 }

        B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

           B = W

In this frame of reference, the variable y'  when it is oscillating is positive and negative, therefore Newton's equation remains

         B’= m \frac{d^2 y'}{dt^2 }

           

the thrust is given by the Archimedes relation

         B = ρ_liquid g V_liquid

     

the volume is

        V = π r² y'

     

we substitute

          - ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

          \frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi  r^2/m ) y' \ =0

this differential equation has a solution of type

         y = A cos (wt + Ф)

where

         w² = ρ_liquid g π r² /m

angular velocity and frequency are related

         w = 2π f

         

we substitute

          4π² f² = ρ_liquid g π r² / m

          f = \frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \  \pi  r^2 \ g}{m } }

calculate

         f = \frac{1}{2 \pi }  \sqrt{ \frac{ 1000 \ \pi  \ 0.03^2 \ 9.8 }{0.025}  }

         f = 5.3 Hz

6 0
3 years ago
a 1 m length of string is wrapped around a solid disk, of mass .25 kg and a radius of .30m, mounted on a frictionless axle. the
Nezavi [6.7K]

Answer:

60 rad/s

Explanation:

∑τ = Iα

Fr = Iα

For a solid disc, I = ½ mr².

Fr = ½ mr² α

α = 2F / (mr)

α = 2 (20 N) / (0.25 kg × 0.30 m)

α = 533.33 rad/s²

The arc length is 1 m, so the angle is:

s = rθ

1 m = 0.30 m θ

θ = 3.33 rad

Use constant acceleration equation to find ω.

ω² = ω₀² + 2αΔθ

ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)

ω = 59.6 rad/s

Rounding to one significant figure, the angular velocity is 60 rad/s.

8 0
3 years ago
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