All categories

3 years ago

What is the speed of an object traveling a distance of 25 meters in 25 seconds

Physics

2 answers:

tiny-mole [99]3 years ago

7 0

Answer: 2.23694 miles per hour

Explanation:

sergeinik [125]3 years ago

6 0

Answer:

1 meter per second

Explanation:

Speed=Distance/Time

S=D/T

S=25/25

S=1

You might be interested in

Which quantity is a scalar quantity?

Vsevolod [243]

The answer is Area ,area

5 0

3 years ago

When a ball increases in speed by the same amount each second its acceleration?

MAXImum [283]

Its acceleration is constant.

5 0

3 years ago

The height a ball bounces is less than the height of the previous bounce due to friction. The heights of the bounces form a geom

AlladinOne [14]

Answer:

Explanation:

Let the first height be h . second height .75h

third height .75h . fourth height .75²h

fifth height .75²h , sixthth height .75³ and so on

Total distance consists of two geometric series as follows

1 ) first series

h + .75h + .75²h + .75³h......

2 ) second series

.75h +.75²h +.75³h + .75⁴h .......

Sum of first series :

first term a = h , commom ratio r = .75

sum = a / (1 - r )

= h / 1 - .75

= h / .25

sum of second series :--

first term a = .75 h , commom ratio r = .75

sum = a / (1 - r )

= .75h / 1 - .75

= .75h / .25

Total of both the series

= 4h + 3h

= 7h .

h = 1 m

Total distance = 7 m

6 0

3 years ago

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$