Ask question

Ask question

All categories

3 years ago

8

On your wedding day, you leave for the church 25 minutes before the ceremony is to begin. The church is 8 miles away. On the way

, you have to make an unanticipated stop for construction work. As a result, your average speed for the first 15 minutes is only 7 miles per hour. What average speed in miles per hour do you need for the rest of the trip to get to the church in time

1 answer:

natulia [17]3 years ago

6 0

Answer:

V=37.5miles/h

Explanation:

For convenience, let's convert the average speed to miles per minute:

V=7miles/h * 1h/60min = 0.1167 miles/min.

The distance traveled during the first 15 min was:

D = V*t = 1.75 miles So, the remaining distance is 6.25miles.

Since you only have 10min left:

Vr = Dr / tr = 6.25 / 10 = 0.625 miles / min. If we take that to miles per hour we get final answer:

Vr = 0.625 miles/min * 60min/1h = 37.5 miles/h

You might be interested in

If the temperature of the conductor is increased, the electrons’ speeds decrease

Tatiana [17]

Answer:

HOPE IT HELPS YOU!!!

Explanation:

Mark FadedGirl25 as brainliest

5 0

3 years ago

Which form of energy does the motor convert most of its electric energy into?

aev [14]

Answer:

Kinetic energy

Explanation:

6 0

2 years ago

Read 2 more answers

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determi

DerKrebs [107]

Answer:

Weight of the woman in Newton

$x = 698.6 \ N$

Mass of the woman in slug

$Mass = 4.86 \ slug$

Mass of the woman in kg

$Mass = 16 \ kg$

My weight in Newton

$W = 784 \ N$

Explanation:

From the question we are told that

The weight of the woman in pounds is $W = 157 \ lb$

Converting to Newton

1 N = 0.22472 lb

x N = 157

=> $x = \frac{157 * 1}{0.22472}$

=> $x = 698.6 \ N$

Obtaining the mass in slug

$Mass = \frac{W}{g}$

Here $g = 32.2 ft/s^2$

So

$Mass = \frac{157 }{32.2}$

$Mass = 4.86 \ slug$

Obtaining the mass in kilogram

$Mass = \frac{W}{g}$

Here $g = 9.8 \ m/s$

So

$Mass = \frac{157 }{9.8}$

$Mass = 16 \ kg$

Generally weight is mathematically represented as

$W = m * g$

Given that my mass is 80 kg then my weight is

$W = 80 *9.8$

$W = 784 \ N$

6 0

2 years ago

What is the measuring devices of a mass

erik [133]

The measuring devices for mass is scales and Balances

6 0

2 years ago