All categories

3 years ago

Whole numbers smaller than 20

Physics

2 answers:

Kruka [31]3 years ago

5 0

{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

Gennadij [26K]3 years ago

3 0

Answer:

19 and below because a whole number is just a number that isn't a fraction. They are basic counting numbers

Explanation:

You might be interested in

Scientists have detected sound waves with frequencies as low as 13 Hz being produced by

Lena [83]

Answer:

The wavelength of the sound waves are, λ = 26.92 m

Explanation:

Given data,

The frequency of the sound waves, f = 13 Hz

The speed of sound waves in air, v = 350 m/s

The velocity of the wave in the medium is equal to the product of its wavelength and its frequency.

The formula for the wavelength of the waves is,

λ = v / f

Substituting the given values in the above equations,

λ = 350 m/s / 13 Hz

= 26.92 m

Hence, the wavelength of the sound waves are, λ = 26.92 m

4 0

4 years ago

A car traveling at an initial velocity of 12.0 m/s accelerates at a constant rate to 20.0 m/s over a time of 4.00 s. How far did

nasty-shy [4]

1 second - 32m
2 second - 84m
3 second - 156m
4 second - 248m
Hope this helps

6 0

3 years ago

Read 2 more answers

A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine

klemol [59]

I think the answer for the question above its b 1.2

6 0

4 years ago

Read 2 more answers

A fighter jet of mass 9600 kg is moving at a speed of 72 m/s. if its total energy is 2.60x10^8 j, how high from the ground is it

Ilia_Sergeevich [38]

Answer:

The correct answer is

Explanation:

3 0

2 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

Firstly we find the vertical component of the initial velocity:

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

Now we find the time taken to ascend to this height:

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

Time taken to descent the total height:

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

Now the total time taken by the ball to hit the ground:

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

Whole numbers smaller than 20 ​

Whole numbers smaller than 20