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3 years ago

Hey people! I sorta need help. :))

Physics

2 answers:

il63 [147K]3 years ago

4 0

Answer:

1. Attract

2. Repel

3. Attract

4 Repel

always remember when dealing with magnets, Opposites attract

Gwar [14]3 years ago

4 0

ATTRACT
REPEL
ATTRACT
REPEL

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Question 1 (1 point)

deff fn [24]

Answer:

Travelled 18 km, they are 6 km from home.

Explanation:

12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.

5 0

3 years ago

A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in ai

Svetach [21]

Answer:

The values is $f_b =14.9 \ beats/s$

Explanation:

From the question we are told that

The speed of the fire engine is $v = 5\ m/s$

The frequency of the tone is $f = 500 \ Hz$

The speed of sound in air is $v_s = 340 \ m/s$

The beat frequency is mathematically represented as

$f_b = f_a - f$

Where $f_a$ is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as

$f_a = [\frac{v_s + v }{v_s -v} ]* f$

substituting values

$f_a = [\frac{340 + 5 }{340 -5} ]* 500$

$f_a = 514.9 \ Hz$

Thus

$f_b =514.9 - 500$

$f_b =14.9 \ beats/s$

8 0

3 years ago

A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.155 µC is suspended from an insulating thread and hangs

adell [148]

Answer:

the magnitude of the charge Q on each plate is $3.053 *10^{-8} \ C$

Explanation:

Given that :

mass (m) = $7.00 *10 ^{-3} \ kg$

charge (q) = +0.155 µC = $+0.155 *10^{-6}\ C$

angle $\theta = 30^0 \ C$

Area A on each plate = 0.0135 m²

From the diagram below;

$tan \ \theta = \frac{Eq}{mg}$ ----- equation (1)

Also by using Gauss Law ;

$Q = \epsilon_0 \phi$

$Q = \epsilon_0EA$ ----- equation (2)

Combination equation 1 and 2 together ; we have

$Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}$

$Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}$

$Q = 3.053 *10^{-8} \ C$

8 0

3 years ago

A flying stationary kite is acted on by a force of 9.8 N downward. The wind exerts a force of 45 N at an angle of 50.0 degrees a

Savatey [412]

Answer:

38 N, 40.0° below the horizontal

Explanation:

Force exerted by an object equals mass times acceleration of that object: F = m ⨉ a. To use this formula, you need to use SI units: Newtons for force, kilograms for mass, and meters per second squared for acceleration.

7 0

3 years ago

The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time

Makovka662 [10]

Explanation:

We have,

Surface area, $A=2\ cm^2=0.0002\ m^2$