Ask question

Ask question

All categories

3 years ago

7

A bird flies 200 km in 4.5 hours. What is the birds average speed?

1 answer:

Mazyrski [523]3 years ago

4 0

900km just multiple

You might be interested in

A runner accelerates to 4.2 m/s2 for 10 seconds before winning the race. How far did he/she run?

timofeeve [1]

She run the distance of 42metres

8 0

3 years ago

Read 2 more answers

The weight of a metal bracelet is measured to be 0.100 N in air and 0.092 N when immersed in water. Find its density

max2010maxim [7]

If you subtract the two weights, 0.100 N and 0.092 N, we will get 0.008. So, the density between 0.100 and 0.092 is 0.008. :)

Answer: 0.008

I hope this helped!:)

4 0

3 years ago

A rocket burns fuel as it shoots into the sky. What happens to the mass and volume of the rocket?

diamong [38]

Answer:

mass goes down volume remains the same

6 0

3 years ago

What do these circuits all have in common

olga55 [171]

They all have a battery.

7 0

4 years ago

Read 2 more answers

A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

Dennis_Churaev [7]

Answer:

The value of x is $x = 0.1955\ m$

The value of the velocity at the top is $v_{top} = 2.25 \ m/s$

Explanation:

The diagram of this process is on the first uploaded image

From the question we are told that

The mass of the block is $m_b = 0.490kg$

The distance of compression is $x$

The force constant is $k = 450 N/m$

The radius of the circular track is $R =1.00m$

The speed at the bottom of the track is $v_A = 13.4 \ m/s$

The frictional force experienced is $F_f = 7.00 \ N$

Now looking at this process we see that the potential energy of the spring is been transformed into the kinetic energy of the block . So,

$PE \ of \ spring = KE \ of \ block$

mathematically i.e

$\frac{1}{2}k x^2 = \frac{1}{2} m_bv^2_A$

$0.5 * 450 * x^2 = 0.5 * 0.490 * 13.4^2$

Making x the subject of the formula

$x = \sqrt{\frac{0.5 * 0.490 * 13.4^2}{ 0.5 * 450} }$

$= 0.1955\ m$

The average workdone by friction is $W_f = F_f * \pi$

Here $\pi is \ the\ net\ displacement$

$W_f = 7 * 3.142$

$= 21.99 J$

The kinetic energy at the bottom is

$KE = \frac{1}{2} * 0.490 * 13.4^2$

$= 43.99 \ N$

The potential energy gained at the top of the circle is

$PE = m_bgh$

Here h is the height which is equal to d(diameter) = 2r = 2 × 1 = 2 m and

g is acceleration due to gravity $= 9.8m/s^2$

Now substituting values

$PE = 0.490 * 9.8 *2$

$=9.604 J$

Since energy is can not be created nor destroyed but transformed according to first law of thermodynamics

$KE_{at \bottom} = PE _ {\ at \ top } + W_f + KE_{\ at \ top}$

$43.99 = 9.604 +21.99 + [\frac{1}{2} m_b * v_{top} ^2]$

$12.369= [2.45* v_{top} ^2]$

$v_{top} =\sqrt{\frac{12.369}{2.45} }$

$= 2.25 \ m/s$

8 0

3 years ago