Ask question

Ask question

All categories

2 years ago

9

What is your average velocity if you drive a distance of 113 km at a speed of 47 km/h, then the same distance at a speed of 66 k

m/h? Answer in units of km/h.

2 answers:

Sauron [17]2 years ago

8 0

Answer:

$V_{avg}=54.90\ km/h$

Explanation:

Given that

Drive 113 km at 47 km/h and another 113 km at 66 km/h.

When distance is same then average velocity can be found by using below formula:

$V_{avg}=\dfrac{2V_1V_2}{V_1+V_2}$

Here

$V_1= 47\ km/h$

$V_2= 66\ km/h$

Now by putting the values

$V_{avg}=\dfrac{2V_1V_2}{V_1+V_2}$

$V_{avg}=\dfrac{2\times 47\times 66}{47+66}$

So

$V_{avg}=54.90\ km/h$

Svetlanka [38]2 years ago

3 0

Answer:

Average velocity v = 54.90 km/h

Explanation:

We have given the distance d = 113 km

First speed = 47 km/h

Time taken to cover 113 km with speed of 47 km/h

$t=\frac{distance}{speed}=\frac{113}{47}=2.4042h$

Time taken to cover 113 km with speed of 66 km/h

$t=\frac{distance}{speed}=\frac{113}{66}=1.7121h$

So total time = 2.4042+1.7121 =4.1163 hour

Total distance = 113+113=226 km

So average velocity $v=\frac{distance}{time}=\frac{226}{4.1163}=54.90km/h$

You might be interested in

If wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:

GenaCL600 [577]

The frequency of the wheel is given by:
$f= \frac{N}{t}$
where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:
$f= \frac{100}{10 s}=10 s^{-1}$

And now we can find the angular speed of the wheel, which is related to the frequency by:
$\omega=2 \pi f=2 \pi (10 s^{-1})=62.8 s^{-1}$

6 0

3 years ago

Which planets are visible without a telescope

Annette [7]

The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.

3 0

2 years ago

A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

6 0

2 years ago

Why is scalar quantity negative charges is not possible

umka2103 [35]

Answer:

Scalar quantity can never be Negative. Because scalar has only magnitude not direction. And magnitude can't be negative.

Explanation:

4 0

2 years ago

Read 2 more answers

The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 9

Ivanshal [37]

Answer:

At light intensity I = 3, is P a maximum

Explanation:

Given:

$P=\frac{90I}{I^2+I+9}$

now differentiating the above equation with respect to Intensity 'I' we get

$\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}$

Now for the maxima $\frac{dP}{dI}=0$

thus,

$0=\frac{-90I^2+810)}{(I^2+I+9)^2}$

or

$-90I^2+810=0$

or

$I^2=\frac{810}{90}$

or

$I^2=9$

or

I = 3

thus, <u>for the value of intensity I = 3, the P is maximum</u>

at I = 3

$P=\frac{90\times3}{3^2+3+9}$

or

$P=\frac{270}{21}$

or

$P=12.85$

5 0

3 years ago