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3 years ago

12

As soon as a bowling ball rolls off the edge of a table its horizontal component of velocity

1 answer:

IgorLugansk [536]3 years ago

7 0

Answer:

B) Remains constant

Explanation:

The motion of the ball is the motion of a projectile, which consists of two separate motions:

- A horizontal motion with constant velocity, since there are no forces acting along the horizontal direction (if we neglect air resistance)

- A vertical motion with constant acceleration, due to the presence of the force of gravity (pointing downward), that accelerates the ball towards the ground with acceleration g=9.8 m/s^2.

We are only interested in the horizontal motion of the ball: we said that there are no forces acting along the horizontal direction, therefore the horizontal component of the the velocity of the ball is constant.

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An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass? show your work

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2 years ago

A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

or

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = $\frac{3RT}{m}$

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = $\frac{3\times8.314\times292}{0.004}$

or

rms speed = 1349.35 m/s

5 0

3 years ago