Ask question

Ask question

All categories

Aleksandr-060686 [28]

3 years ago

5

Distanţa Buzău - Iaşi este de 72 km . Aflaţi timpul în care un autovehicul parcurge această distanţă, dacă are viteza de 72 km p

e oră. Exprimă viteza în unitatea de măsură în SI. Dau coroana !Multumessccc

1 answer:

GrogVix [38]3 years ago

3 0

The distance from Buzău to Iaşi is 72 km. Find out the time a vehicle travels this distance if it has a speed of 72 km per hour. Expresses the speed in the unit of measurement in SI. I give the crown! Thank you

Answer:

3600seconds

Explanation:

Given parameters:

Distance = 72km

Speed = 72km/hr

Unknown:

Time it takes for the vehicle to travel the distance = ?

Solution:

Speed is the distance divided by time;

Speed = $\frac{distance}{time}$

time = $\frac{distance}{speed }$

Insert the parameters;

time = $\frac{72}{72}$ = 1hr

The SI unit of time is in seconds;

3600 seconds is 1hr

The time is 3600seconds

You might be interested in

A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

$T=2.1371sec$

8 0

3 years ago

n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us

barxatty [35]

Answer:

The answer is " $(1.265 \pm 0.010) \ s \ and \ 0.709 \%$ "

Explanation:

In point i:

$T_{theo}= 2\pi \sqrt{\frac{l}{g}}$

$=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s$

If error in the theoretical time period :

$\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}$

$= 0.010 \ s$

$T_{theo} = (1.265 \pm 0.010) \ s$

In point ii:

$\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100$

<h3> $= \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%$ </h3>

5 0

2 years ago

PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE

luda_lava [24]

Australia separated from other continents and species there evolved independently

8 0

3 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

3 years ago

Which of the following modifications to a solenoid would be most likely to decrease the strength of its magnetic field?

Goryan [66]

By reading the fine details of the question, carefully and analytically, I have determined that there's no list of modifications to choose from.

The strength of the magnetic field of a solenoid depends on the electric current in its coil windings, the number of wire turns in its coil windings, and the material in its core.

In order to <em>DE</em>crease the strength of its magnetic field, any one or more of these steps could do the job:

-- DEcrease the electric current in its coil windings. This can be accomplished by decreasing the voltage of the power source that energizes the coil, and/or increasing the resistance of the wire in the coil.

-- DEcrease the number of wire turns in the coil.

-- If the solenoid has anything in its core, change the core to something with a lower magnetic 'permeability'. An Iron core will produce the greatest magnetic field strength. Air, vacuum, or NO core will produce the lowest magnetic field strength.

8 0

3 years ago