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3 years ago

An object’s momentum includes its __________.

Physics

1 answer:

sveta [45]3 years ago

7 0

Answer: Momentum can be defined as "mass in motion

Explanation:

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A ball is revolving horizontally in a circle of radius 1.8 m, and is held by a rigid, massless rod. The mass of the ball is 0.1

OLga [1]

Answer:

$\omega=15 \ rad/s$

Explanation:

given,

radius of circle = 1.8 m

mass of the ball = 0.1 Kg

linear speed of the ball = 27 m/s

angular velocity of orbit = ?

$v = r \omega$

ω is the angular speed of the circle

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{27}{1.8}$

$\omega=15 \ rad/s$

the angular velocity of the orbit is $\omega=15 \ rad/s$

4 0

3 years ago

The open-circuit voltage of a car battery is measured to be 12 V. During engine startup, the battery delivers 10 A to the starte

dmitriy555 [2]

Answer:

- the Thevenin equivalent circuit for the battery is uploaded below

- the battery delivered 117 watts of power to the starter motor

Explanation:

Given the data in the equation

diagram of the Thevenin equivalent circuit for the battery is uploaded below.

Current I = 10 A

Voltage 1 = 12 V

voltage 2 = 11.7 v

R = (V1 - V2) / I

R = (12-11.7)/10

R = 0.3 / 10

R = 0.03Ω

Thevenin equivalent circuit

$R_{L}$ = V2 / I = 11.7 / 10

$R_{L}$ = 1.17Ω

so, power delivered to the starter motor will be;

p = (V2)² / $R_{L}$

P = ( 11.7 V )² / 1.17Ω

p = 136.89 / 1.17

p = 117 watts

Therefore, the battery delivered 117 watts of power to the starter motor

8 0

3 years ago

Size relationships between parts of a whole are known as.

Stella [2.4K]

Answer:

proportion

Explanation:

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8 0

2 years ago

How much voltage is required to run 0.054 A of current through a 15 resistor? Use AV=IR. O A. 2.8 x 102 v O B. 1.2 V O C. 3.6 x

olga55 [171]

Answer:the answer is D

Explanation:

Plug variables in and calculate

6 0

3 years ago

Read 2 more answers

A solid spherical conductor has a radius of 12 cm. The electric field at 24 from the center of the sphere has a magnitude of 640

hoa [83]

Answer:

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

Explanation:

Given:

Radius of sphere (r) = 12 cm = 0.12 m

Distance from the electric field R = 24 cm = 0.24 m

Magnitude (E) = 640 N/C

Find:

Charge density on the sphere

Computation:

Charge on the sphere (q) = (1/K)ER² (K = 9 × 10⁹)

Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²

Charge on the sphere (q) = 4 × 10⁻⁹ C

Charge density on the sphere = q / [4πr²]

Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]

Charge density on the sphere = [4 × 10⁻⁹] / [0.18]

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

4 0

3 years ago