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3 years ago

5

Lúc 7g bạn an đi từ nhà đến trường với tóc độ trung bình là 20km/h . Bạn đến trường lúc 7g20. Tính khoảng cách từ nhà tới trường

?

1 answer:

Len [333]3 years ago

7 0

Answer:

Distance = 6.667 kilometres

Explanation:

Given the following data;

Speed = 20 km/h

Departure time = 7:00

Arrival time = 7:20

Time taken = 20 minutes

To calculate the distance travelled from home to school;

First of all, we would have to convert the value of time in minutes to hours.

Conversion:

60 minutes = 1 hour

20 minutes = X hours

Cross-multiplying, we have;

X = 20/60 = 1/3 hours

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 20 * 1/3

Distance = 20/3 =

Distance = 6.667 kilometres

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A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 15.6

VladimirAG [237]

To solve this problem it is necessary to take into account the kinematic equations of motion and the change that exists in the volume flow.

By definition the change in speed is given by

$v_f^2-v_i^2 = 2ax$

Where,

x= distance

$v_f =$ final velocity

$v_i =$ initial velocity

a = acceleration

On the other hand we know that the flow of a fluid is given by

$\dot{V} = Av$

Where,

A = Area

v = Velocity

PART A )

Applying this equation to the previously given values we have to

$v_f^2-v_i^2 = 2ax$

$v_f^2-0 = 2*(9.8)(15.6)$

$v_f^2=305.76$

$v_f = 17.48$

Therefore the velocity of the water leaving the hole is 17.48m/s

PART B )

In the case of the hole we take the area of a circle, therefore replacing in the flow equation we have to,

$\dot{V} = \pi r^2 v$

$r = \sqrt{\frac{\dot{V}}{\pi v}}$

$r = \sqrt{\frac{3*10^{-3}*\frac{1}{60}}{\pi (17.48)}$

$r = \sqrt{9.10*10^{-7}}$

$r = 0.54*10^{-4}$

The diameter is 2 times the radius, then is $1.91*10^{-3}$ m or 1.91mm

<em>Note: The rate flow was converted from minutes to seconds.</em>

8 0

3 years ago

From measurements on a uniformly sized material from a dryer, it was found that the surface area of the material is 1200 m2. If

amm1812

Answer:

$D = 2.07\times 10^{-4} m$

Explanation:

We know that the volume of the material

= mass/density

mass= 360 Kg

density = 1450 kg/m^3

Volume= 360/1450= 0.248 m^3

and Volume = Area× equivalent Diameter

Area= 1200 m^2

⇒Equivalent Diameter= Volume/Area= 0.248/1200=

$D = 2.07\times 10^{-4} m$

3 0

3 years ago

Am object of ma 4 i thrown vertically upward from level ground with initial peed of 20. How high did the object go

denis23 [38]

The maximum height of the object when thrown at the speed of 20 m/s is 20 m.

What is speed?

Speed is the magnitude of the rate of change of the position of any object with time which is a scalar quantity. It is the ratio of speed by time with the unit m/s or cm/s.

The height this object can travel with initial speed of 20 m/s can be calculated as,

<u>v^2 = u^2 + 2as</u>

where v is the final velocity, u is the initial velocity a is the acceleration due to gravity s is height.

0 = 20^2 + 2 (-10) s

s = 400 / 20 = 20 m

Therefore, the maximum height the object can travel with initial speed of 20 m/s is 20m.

To learn more about the speed click on the given link brainly.com/question/13943409

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8 0

2 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

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3 years ago

What information do you need to describe an object's location

stira [4]

longitude and latitude<span />

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3 years ago