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3 years ago

I NEED HELP ANSWER FAST WILL GIVE BRAINLY!!!!!!!

Physics

2 answers:

trapecia [35]3 years ago

7 0

Answer:

the number of protons.

Explanation:

the atomic number is the number of protons in the nucleus of an atom. the number of protons Define the identity of an element.

MAVERICK [17]3 years ago

5 0

Answer:

The number of protons in one atom of an element determines the atom's identity, and the number of electrons determines its electrical charge. The atomic number tells you the number of protons in one atom of an element. It also tells you the number of electrons in a neutral atom of that element.

Explanation:

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"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th

Semenov [28]

Answer:

$f_L = 1000 Hz$

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

$f_L = \dfrac{v+v_L}{v+v_s}f_s$

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

v_s is the speed of source

now,

considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

$f_L = \dfrac{v+14}{v+14}\times 1000$

now on solving we will get

$f_L = 1000 Hz$

hence, the frequency heard by the listener is equal to 1000 Hz

7 0

3 years ago

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th

il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).

The amount of weight displaced or the amount of buoyant force is:

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

average density = 1.07 kg/L = 1.07 g/mL

5 0

3 years ago

Read 2 more answers

A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

$\omega = 2 \pi f$ (1)

where

$\omega$ is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$ , the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz$

And the frequency is the number of complete revolutions made per second.

For an object in circular motion, the tangential speed is related to the angular speed by the equation

$v=\omega r$

where

$\omega$ is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

$d=vt$