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2 years ago

Which of the following statements apply to electric charges? please help this is the only question i need done!!

2 answers:

8 0

The only correct statement on that list is <em>choice-C</em>: If a positively charged rod is brought close to a positively charged object, the two objects will repel.

nekit [7.7K]2 years ago

5 0

Answer:

C. If a positively charged rod is brought close to a positively charged object, the two objects will repel.

Explanation:

According to coulomb's law, the magnitude of the electric force between two point charges at rest is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them. This force is repulsive for like charges and attractive for opposite charges.

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Help please :pensive:

tino4ka555 [31]

Answer:

0m/s²

Explanation:

Given parameters:

Initial velocity of the boat = 8m/s

Final velocity = 8m/s

Time taken = 4s

Unknown:

Acceleration of the boat = ?

Solution:

Acceleration is the rate of change of velocity with time.

It is mathematically expressed as;

A = $\frac{v - u}{t}$

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

Insert the parameters and solve;

A = $\frac{8-8}{4}$ = 0m/s²

6 0

2 years ago

Which of the following devices is associated with the reception of radio signals?

ki77a [65]

What are the "following" devices ? ?
I think they're a list of choices that you have but aren't sharing.

A few devices associated with the reception of various types of
radio signals include the resonant tank, the local oscillator, the
mixer, the detector, the coherer, the discriminator, the parabolic
reflector, the lecher wires, the audio transducer, the demultiplexer,
and ... my personal guess ... the 'antenna' or 'aerial'.

6 0

3 years ago

Read 2 more answers

A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m

Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

Horizontal displacement = 40.5 m

Time

$t=\frac{40.5}{28.64}=1.41s$

Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

Time to reach the goal posts 40.5 m away = 1.41 seconds

Acceleration = -9.81m/s²

Substituting in s = ut + 0.5at²

s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

Height of throw = 1.4 m

Height traveled by ball = 2.83 m

Total height = 2.83 + 1.4 = 4.23 m

When the ball goes to first base it will be 4.23 m high.

8 0

3 years ago

Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius

Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

$a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2$

The centripetal acceleration for the second radius; 4.0 m

$a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2$

The centripetal acceleration for the third radius; 6.0 m

$a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2$

The centripetal acceleration for the fourth radius; 8.0 m

$a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2$

The centripetal acceleration for the fifth radius; 10.0 m

$a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2$

6 0

3 years ago

A yellow train of mass 100 kg is moving at 8 m/s towards an orange train of mass 200 kg traveling on the opposite direction on t

vladimir1956 [14]

Mass of yellow train, my = 100 kg

Initial Velocity of yellow train, = 8 m/s

mass of orange train = 200 kg

Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)

To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which

The sum of initial momentum = the sum of final momentum

Since the question only wants the sum of initial momentum,

(100)(8) + (200)(-1) = 600 m/s

8 0

2 years ago