تقطع اولا مسافة 8 km شمالا من البيت ثم تمشي شرقا حتى تكون ازاحتك من البيت 10km ما مقدار المسافة التي قطعتها شرقا

Kinematics

leonid [27]

Answer:

a)

a = 2 [m/s^2]

b)

a = 1.6 [m/s^2]

c)

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

a)

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

$v_{f} = v_{i}+(a*t)$

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

40 = 0 + (a*20)

a = 2 [m/s^2]

The distance can be calculates as follows:

$v_{f} ^{2} = v_{i} ^{2}+(2*a*x)$

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

where:

x2 = distance [m]

t2 = 30 [s]

x2 = 40*30

x2 = 1200 [m]

b)

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

$v_{f} = v_{i}-a*t$

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

$v_{f} ^{2} = v_{i} ^{2} -2*a*x3\\$

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

c)

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

8 0

3 years ago

Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

For the second mirror will be the same value

$A = A_{mirror1}+A_{mirror2}$

$A = 55.41m+55.41m$

$A= 110.82m$

Case for Keck 1 telescope:

$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

Case for Hobby-Ebberly telescope:

$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

Case for Subaru telescope:

$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

Case for Gemini North telescope:

$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

Key term:

Photons: particles that constitute light. 

3 0

3 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$