Suppose a new asteroid was recently discovered which takes 557 months to orbit the Sun once (that's equal to 16,700 days or 46.4

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Diameter of the smaller paintball = 5 cm

d₂ = Diameter of the larger paintball = 9 cm

V₂ = Volume of the larger paintball

Volume of the smaller paintball

Likewise

By dividing the previous two equations, we derive

∴ The larger one accommodates 163.296 cm³ of paint

Response:

Clarification:

Typically, the angle between the y-axis and the x-axis measures 90°, and it is understood that for maximum distance, the angle must be 45° relative to the horizontal. Mo should release the ball approximately halfway between a straight throw in front and a vertical throw.

Answer:

The question is incomplete; refer to the attachment for the diagram

Explanation:

Provided that,

The mass of the car

M = 1500kg

Entering the curve at Point A with a speed of

Va = 100km/hr = 100× 1000/3600

Va = 27.78m/s

The car slows down at a constant rate until it reaches point C at a speed of

Vc = 50km/hr = 50×1000/3600

Vc = 13.89m/s

Radius of curvature at point A

p = 400m

Radius of curvature at point B

p = 80m

The distance from point A to point B as indicated in the attachment is

S=200m

We need to determine the total horizontal forces at points A, B, and C applied by the road on the tire

The constant tangential acceleration can be computed using the equation of motion

Vc² = Va² + 2as

13.89² = 27.78² + 2 × a × 200

192.9 = 771.6 + 400a

400a = 192.9—771.6

400a = -578.7

a = -578.7 / 400

a = —1.45 m/s²

at = —1.45m/s²

The tangential acceleration is -1.45m/s² and it is negative due to the car decelerating

Since the vehicle is decreasing speed at a uniform rate, the tangential acceleration remains the same at every point

At point A

at = -1.45m/s²

At point B

at = -1.45m/s²

At point C

at = -1.45m/s²

Now,

We can compute the normal component of acceleration (centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration can be calculated using

ac = v²/ p

At point A ( p = 400)

an = Va²/p = 27.78² / 400

an = 1.93 m/s²

At point B (p = ∞), since point B is the inflection point

Then,

an = Vb²/p =  Vb/∞ = 0

an = 0

At point C ( p = 80m)

an = Vc²/p = 13.89² / 80 = 2.41m/s²

an = 2.41 m/s²

Then,

The tangential force is

Ft = M•at

Ft = 1500 × 1.45

Ft = 2175 N.

Given that the tangential acceleration remains constant, this represents the tangential force at A, B, and C

Next, the normal force

At point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N.

Thus, the horizontal force acting at each point is

Using the right triangle vector method

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N

Answer:

The positioning of the object along the principal axis relative to the concave mirror.

Explanation:

In a concave mirror, the characteristics of the image generated depend on where the object is situated in relation to the mirror. The distance from the mirror to the object positioned along the principal axis is key.

The nearer the object is to the mirror, the larger or more magnified the image will appear. For example, placing an object between the focal point and the concave mirror's pole results in a significantly larger image compared to an object placed outside the center of curvature of the mirror.

I presume that the engine operates under a Carnot cycle, which allows for maximum efficiency.

Assuming this, the Carnot cycle's efficiency can be expressed as

where
represents the temperature of the cold sink
and

signifies the temperature of the heat source.
In this scenario, the cold sink temperature is 313 K, while the heat source temperature is 425 K; thus, we can calculate the engine's efficiency as
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