Ask question

Ask question

All categories

3 years ago

14

4. How long will it take a car travelling with a speed of 160 km hr to cover a distance of 700 meters? Hint: km/hr should be con

verted to m/s

1 answer:

Inessa [10]3 years ago

3 0

Answer:

15.8 seconds

Explanation:

Create an extended calculation to convert all the unit to what you need.

160 km 1000 m 1 hour 1 min

----------- x ------------- x -------------- x ---------- = 44.4 m/s

1 hour 1 km 60 min 60 sec

So 160km/hr is equal to 44.4m/s

Now you can figure out how many seconds it will take to go 700 meters.

44.4 m

---------- X x sec = 700 m

1 sec

Solve for x sec

x sec = 700m / 44.4 m/s

= 15.8 seconds

You might be interested in

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago

A car's fuel economy is 30 mpg. a warning light is displayed if the remaining distance that can be driven before the car runs ou

Alchen [17]

So 10 gallons of gas would let you travel 300 Miles.

x gallons = 50 Miles

10 : 300 :: x : 50

x = 500/300

x = 1.66667 gallons.

So, the car would run 10 - 1.6666 gallons = 8.33 gallons.

After that, the warning light turns ON!

Hope this helps!!

7 0

3 years ago

Read 2 more answers

The surface of the Earth changes from processe such as erosion. Which of these changes to Earth's surface is an example of erosi

Ad libitum [116K]

Answer:

D. the wind picking up dust and carrying it

Explanation:

Erosion is a process in which an agent transfer the top soil to another region, thereby exposing the lower soil. These agents have the ability to move the top layer of soil and deposit it at another place. The major agents in this case are; a running or flowing body of water and wind.

Therefore, the change to the Earth's surface that is an example of erosion is the wind picking up dust and carrying it. Thereby exposing the lower layers.

6 0

2 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago

My sis needs help with this and I don't wanna help her!!! Can you guys help her????

Katena32 [7]

Answer and Explanation:

Cup 1. The ice cubes are in a solid state.

Cup 2: The cup has water that is liquified.

Cup 3: The cup is empty and only contains free floating air particles.

***If you found my answer helpful, please give me the brainliest. :) ***

7 0

3 years ago

Read 2 more answers