Which two biomes have little precipitation?

Where is heat transferred by conduction in a lava lamp?

Semmy [17]

It is transferred by direct contact, say you touched it, you would feel the heat

8 0

3 years ago

A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai

Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a) Let us assume the depth be h

As we know that

$Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h$

After solving this,

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that

height of the extra fluid is

$h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}$

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

$Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000$

Pb' = 122 KPa

3 0

3 years ago

A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is

Anika [276]

Answer:

The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.

Explanation:

For 2 quantities A and B represented as

$A\pm \Delta A$ and $B\pm \Delta B$

The sum is represented as

$Sum=(A+B)\pm (\Delta A+\Delta B)$

For the the values given to us the sum is calculated as

$Sum=(2.9+3.9)\pm (0.1+0.2)$

$Sum=6.8\pm 0.3$

Now the since the uncertainity inthe sum is $\pm 0.3$

The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity

Thus closest distance equals $6.8-0.3=6.5$ meters

3 0

3 years ago

You want to manufacture a guitar such that the instrument will be in tune when each of the strings are tightened to the same ten

charle [14.2K]

Answer:

0.000507 kg/m

Explanation:

L = Length of string

T = Tension

$\mu$ = Mass density of string

E denotes the E string

D denotes the D String

Frequency is given by

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$

So

$f\propto \sqrt{\dfrac{1}{\mu}}$

$\dfrac{f_D}{f_E}=\sqrt{\dfrac{\mu_E}{\mu_D}}\\\Rightarrow \mu_E=\dfrac{f_D^2}{f_E^2}\mu_D\\\Rightarrow \mu_E=\dfrac{146.83^2}{329.63^2}\times 0.00256\\\Rightarrow \mu_E=0.000507\ kg/m$

The mass density of the E string is 0.000507 kg/m

8 0

3 years ago

A block rests on a frictionless table on Earth. After a 40-N horizontal force is applied to the block, it accelerates at 9.2 m/s

Paul [167]

Answer:

4.6 $\frac{m}{s^2}$

Explanation:

Since the table is frictionless, there is no force of dynamic friction between table an block when the horizontal force is applied to it on Earth. Exactly the same is true when the table is taken to the Moon. Therefore, the Net Force acting on the object in both cases when the object accelerates, is the external horizontal force.

Notice that on Earth and on the Moon, the weight of the object (vertical and pointing up) is compensated by the normal force of the table on the object (pointing up and of the same magnitude as the weight) that precludes movement in the vertical direction. So in both cases, its acceleration will only be due to the horizontal force.

We use the equation for Net Force to find the mass of the object:

$F=m*a\\40 N =m * 9.2 \frac{m}{s^2}\\\frac{40}{9.2} kg=m\\m=\frac{40}{9.2} kg$

We use this mass (since the mass of the object is a constant independent of where the object is) to find the acceleration the object will experience when the 20 N horizontal force is applied on it on the Moon:

$f=m*a\\20N=\frac{40}{9.2} kg*a\\20*9.2=40*a\\\frac{20*9.2}{40} =a\\a=\frac{9.2}{2} \frac{m}{s^2} \\ a=4.6 \frac{m}{s^2}$

3 0

3 years ago