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2 years ago

Which TWO statements describe behaviors of particles that are related to the

1 answer:

7 0

Answer: 1: (A) They allow electrons to move freely between them. 2: (C) they change their positions relative to one another.

Explanation:

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Jack travelled 360 km at an average speed of 80 km/h. Elaine

diamong [38]

Answer:

Average speed of Elain = 60 km/h

Explanation:

Total Distance covered by Jack = 360km

Average Speed of Jack = 80 km/h

Time taken by Jack to complete his journey = Distance / Average speed = 360 km / 80 km/h

Time taken by Jack to complete his journey = 4.5 hours

As it is given the both Jack and Elain travelled the same amount of distance:

Total distance travelled by Elain = 360 km

It is given that Elain took 1.5 hourse more than Jack to cover the distance, so Time taken by Elain to cover the distance is = 4.5 hours + 1.5 hours = 6 hours

Average speed of Elain = Distance/ time = 360 km / 6 hours

Average speed of Elain = 60 km/h

3 0

3 years ago

How much watt is called 1 H. P

agasfer [191]

Answer:

746 watts

Explanation:

7 0

2 years ago

A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes.

zaharov [31]

Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m$

$\lambda=8.67\ mm$

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

3 0

3 years ago

Answer the question with step.

Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0

3 years ago

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

2 years ago