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3 years ago

A train, initially moving at 12 m/s, speeds on to 36 m/s in 120 seconds. What is its acceleration?

Physics

1 answer:

Levart [38]3 years ago

6 0

Answer:

0.2 m/s²

Explanation:

acceleration formula is given by the difference in velocities over the time

hence final velocity - intial velocity / time

= 36-12/120

=0.2m/s²

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Radar uses radio waves of a wavelength of 2.4 ${\rm m}$ . The time interval for one radiation pulse is 100 times larger than t

blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x $10^{8}$ m/s)

f= 3 x $10^{8}$ / 2.4

f=1.25 x $10^{8}$ hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x $10^{8}$

T= 8 x $10^{-9}$ s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x $10^{-9}$ => 800 x $10^{-9}$ s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

$x_{min}$ = τc/2 => (800 x $10^{-9}$ x 3 x $10^{8}$ )/2

$x_{min}$ =120m

8 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago

A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c

den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, $L_1=L-1$

New time period of the pendulum is, $T_1=2.81\ s$

We know that, the time period of a simple pendulum of length 'L' is given as:

$T=2\pi\sqrt{\frac{L}{g}}$ -------------- (1)

So, for the new length, the time period is given as:

$T_1=2\pi\sqrt{\frac{L_1}{g}}$ ------------ (2)

Squaring both the equations and then dividing them, we get:

$\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1$

Now, plug in the given values and calculate 'L'. This gives,

$L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m$

Therefore, the original length of the simple pendulum is 2.97 m

4 0

3 years ago

Người ta đúc một khối nhóm (A) và một khối sắt ( (Fe) hình khối với cũng khói lượng 5 kg Khối lượng riêng (2) của Al và Fe là 2

IrinaVladis [17]

Translate to English please

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3 years ago

Which lists three organic biological molecules?

igor_vitrenko [27]

Answer:

Explanation:

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3 years ago