What two things affect the density of Water?

A soccer coach riding his bike reaches his office in xx hours. If he travels at 24 km/h, he reaches his office 5 minutes late. I

Elina [12.6K]

Answer:

18 km

Explanation:

Let 'd' be the distance between his house and office.

Normal time taken to reach the office = 'x' hours.

If speed is 24 km/h, time is increased by 5 minutes.

If speed is 30 km/h, time is reduced by 4 minutes.

We know that,

Time taken = Distance traveled ÷ Speed

So, when speed is 24 km/hr, time is increased by 5 minutes.

$1\ min = \frac{1}{60}\ h\\5\ min =\frac{5}{60}=\frac{1}{12}\ h$

So, time is $x+\frac{1}{12}$

Therefore,

$x+\frac{1}{12}=\frac{d}{24}\\\\x=\frac{d}{24}-\frac{1}{12}\\\\x=\frac{1}{12}(\frac{d}{2}-1)---------1$

Now, when speed is 30 km/h, time is reduced by 4 minutes or $\frac{4}{60}=\frac{1}{15}\ hours$

So, time now is $x-\frac{1}{15}$

Again using the time formula, we have

$x-\frac{1}{15}=\frac{d}{30}\\\\x=\frac{d}{30}+\frac{1}{15}\\\\x=\frac{1}{15}(\frac{d}{2}+1)-------------2$

Equations (1) and (2) are equal. So,

$\frac{1}{12}(\frac{d}{2}-1)=\frac{1}{15}(\frac{d}{2}+1)\\\\\frac{15}{12}(\frac{d}{2}-1)=\frac{d}{2}+1\\\\\frac{5d}{8}-\frac{5}{4}=\frac{d}{2}+1\\\\\frac{5d}{8}-\frac{d}{2}=1+\frac{5}{4}\\\\\frac{5d-4d}{8}=\frac{4+5}{4}\\\\\frac{d}{8}=\frac{9}{4}\\\\d=\frac{9\times 8}{4}=\frac{72}{4}=18\ km$

Therefore, the office is 18 km from his house.

6 0

3 years ago

A 2.31 kg rope is stretched between supports 10.4 m apart. If one end of the rope is tweaked, how long will it take for the resu

zlopas [31]

Answer:

t = 0.657 s

Explanation:

First, let's use the appropiate equations to solve this:

V = √T/u

This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.

Where:

V: Speed of the disturbance

T: Tension of the rope

u: linear density of the rope.

The density of the rope can be calculated using the following expression:

u = M/L

Where:

M: mass of the rope

L: Length of the rope.

We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:

u = 2.31 / 10.4 = 0.222 kg/m

Now, replacing in the first equation:

V = √55.7/0.222 = √250.9

V = 15.84 m/s

Finally the time can be calculated with the following expression:

V = L/t ----> t = L/V

Replacing:

t = 10.4 / 15.84

t = 0.657 s

4 0

4 years ago

A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?

pav-90 [236]

F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs

4 0

3 years ago

Read 2 more answers

A force of 150N is inclined at 50° t The horizontal direction. Find it's components in horizontal and vertical direction

Komok [63]

The solution is attached below.

8 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constan

Fantom [35]

Answer:

(a). The emf induced in the loop is 0.005467 V.

(b). The direction of the current is clock wise direction.

Explanation:

Given that,

Circumference = 165.0 cm

Constant rate = 12.0 cm/s

Magnetic field = 0.500 T

Time = 9.0 s

We need to calculate the emf induced

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA\cos\theta)}{dt}$

$\epsilon=B\dfrac{dA}{dt}$ ...(I)

Area of the coil is

$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$

Put the value of area in equation (I)

$\epsilon=B\times2\pi r\dfrac{dr}{dt}$ ...(II)

The circumference of the coil is

$C=2\pi r$

$r = \dfrac{C}{2\pi}$

Put the value into the formula

$r=\dfrac{165.0}{2\pi}$

$r=26.26\ cm$

The rate of change of radius is

$\dfrac{dr}{dt}=\dfrac{12}{2\pi}$

$\dfrac{dr}{dt}=1.90\ cm/s$

The radius of the coil in 9 sec

$r=26.26-1.90\times9$

$r=9.16\ cm$

Put the value in the equation (I)

$\epsilon=-0.500\times2\pi\times9.16\times10^{-2}\times1.90\times10^{-2}$

$\epsilon=-0.005467\ V$

The magnitude of the emf

$|\epsilon|=0.005467\ V$

(b). Right hand rule :

According to right hand rule,

The thumb shows the direction of force, the index finger shows the direction of magnetic field and the middle finger shows the direction of current.

So, The direction of the current is clock wise direction.

Hence, (a). The emf induced in the loop is 0.005467 V.

(b). The direction of the current is clock wise direction.

8 0

3 years ago