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KengaRu [80]
3 years ago
10

A ball is rolled along the inside of a partial hoop lying flat on a table. which path will the ball follow when it gets to the e

nd of the hoop?

Physics
2 answers:
wlad13 [49]3 years ago
6 0

Explanation :

A ball is rolled along the inside of a partial hoop lying flat on a table. The ball will move in a straight path as shown in the attached figure.

The figure shows a semicircular hoop. When the ball reaches at the other end, it will move tangentially.

In a circular path, the object move with constant speed but the velocity changes at every instant.

So, the ball will move in a straight line when it reaches at the end of the hoop.

Zinaida [17]3 years ago
3 0
Once it leaves the circle, it'll continue in a straight line, in the direction it was moving when it left the circle.
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Which is true about light and heat?
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My best guess is c) Dark colors reflect less radiation making them warmer.

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An electron in an unknown energy level of a hydrogen atom transitions to the n=2 level and emits a photon with wavelength 410 nm
lyudmila [28]

Answer:

The initial energy level = 6

Explanation:

Photon wavelength is proportional to energy. The wavelength of emitted photons is related to the energy levels of the atom as given by the Rydberg formula:

ₕ₁₂

(1/λ) = Rₕ [(1/n₂²) − (1/n₁²)]

where n₂ = final energy level = 2

n₁ = initial energy level = ?

Rₕ = Rydberg's constant = 1.097 × 10⁷ m⁻¹

λ = wavelength = 410 nm = 410 × 10⁻⁹ m

1/(410 × 10⁻⁹) = (1.097 × 10⁷) [(1/2²) − (1/n₁²)]

0.223 = [(1/4) − (1/n₁²)]

(1/n₁²) = 0.02778

n₁² = 1/0.02778 = 36

n₁ = 6.

7 0
3 years ago
A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg
ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

p=m/V\\\\ p=m/(L*A)\\p*A=m/L

Now the equation for speed of traverse wave is calculated through:

\sqrt \frac{F*L}{m}\\

=\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}

Substituting the values

\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3}  \\

=22.49 m/s

4 0
3 years ago
A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3
Kamila [148]

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

7 0
2 years ago
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