A car covers a distance of 200m. If its velocity is 20 m/s, calculate the time taken.

In still water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles downstream,

Vladimir79 [104]

<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16

Traveling against the current:
(18 - c)*t = 8

Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16

Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16

(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0

Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c

So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3

(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3

And it's verified.</span>

4 0

3 years ago

Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured

kvv77 [185]

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

Mass of the machine, m=70 kg
Stiffness of the system, k=30000 N/m
Damping ratio=0.2
Damping force, F=450 N
Angular velocity $\omega=13\ \rm rad/s$

a)We know that the amplitude X at steady state is given by

$X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}\\$

Where

$\omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s$

$\omega=13\ \rm rad/s$
$\rho=0.2$

$F_0=450\ \rm N$
$m=70\ \rm kg$

$X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m$

b) The phase shift of the motion is given by

$\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393\\$

c)Transmissibility ratio is given by

$T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57$

d)The magnitude of the force transmitted to the ground is

$F_T=(T.R)\times F_0\\\\=450\times1.57\\\\=707.6\ \rm N$

e)The maximum velocity is given by $V_{max}$

$V_{max}=\omega A_0\\\\=13\times 0.0229\\\\=0.298\ \rm m/s$

6 0

3 years ago

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a

sertanlavr [38]

Answer:

Electric field, E = $6.36\times 10^{6}\ V/m$

Explanation:

Given that,

The potential difference across the membrane, $\Delta V=0.063\ V$

The thickness of the membrane is, $\Delta x=9.9\times 10^{-9}\ m$

We need to find the magnitude of the electric field in the membrane. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{-\Delta V}{\Delta x}\\\\E=\dfrac{0.063\ V}{9.9\times 10^{-9}\ m}\\\\E=6.36\times 10^{6}\ V/m$

So, the magnitude of the electric field in the membrane is $6.36\times 10^{6}\ V/m$ .

7 0

4 years ago

How does the sun impact land at different latitudes?

Masja [62]

The radiation of the solar rays of the sun can damage and heat up land closer but at the same time cool because of the atmosphere

5 0

3 years ago

Define habitat preservation and give at least two examples.

lisov135 [29]

Habitat preservation is when people get together or just one person helps to stop the end of a living space for animals such as the monterey aquarium the help sea life by doing research to see why animals are dying and do what they can to stop it. another one if nation parks or forests you are not allowed to harm them in any way they are protected by laws(the government)

7 0

3 years ago

A car covers a distance of 200m. If its velocity is 20 m/s, calculate the time taken. ​

A car covers a distance of 200m. If its velocity is 20 m/s, calculate the time taken.