The rest deltoid row is a back exercise true or false

If you had 8 balls and 7 of them were a certain weight, and 1 of them was heavier, how could you find the heaviest ball. All the

dybincka [34]

Answer:

There are two method of comparing the balls 1) using a balance 2) by only 2 weighings.

Explanation:

There are two method of comparing the balls 1) using a balance 2) by only 2 weighings.

Make the following groups - --- (1,2,3),(4,5,6),(7,8)

Step 1. compare the Weigh (1,2,3) and (4,5,6)

there are 2 possible outcomes:

1---both the group are of same weight. and named as (Case A)

2--- one of the group is heavier than other and named as (Case B)

Step 2. Let examine both case

In Case A --in this case, now compare the weight of 7th and 8th ball. By this you have recognize the heavier ball by 2 weighing method.

In Case B -- considered the heaviest group (assume group (1,2,3) is heavy), from this group take randomly two ball and compare the their weight. out of these two ball, one is heavy else the third ball is.

7 0

3 years ago

If the baseball and the plastic ball were moving at the same speed which ball would hit a bat harder

Lerok [7]

A baseball would hit the bat harder. This is because the baseball is a lot heavier and more dense than the plastic ball. The keyword that you're looking for is density. The baseball is dense.

7 0

3 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

3 years ago

PLEASE HELP ASAP! Any absurd answers will be reported. NO LINKS, IDC IF ITS THE ONLY WAY TO GET AN ANSWER.

svetoff [14.1K]

Answer:

There is a thing called a continental drift. It started about 200 million years ago. At first the continents were all attached, this super continent was called pangaea. Continental drift occurs because of the shift of the tectonic plates within the earth's outer shell. The heat from within the earth triggers movement to occur. This a very slow process though. It took 200 million years for the continents to get where the are now and would probably take another 200 to collide.

6 0

3 years ago

Read 2 more answers

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

11111nata11111 [884]

Answer:

$1.62\times 10^{-8}\ \text{s}$

Explanation:

$\epsilon_0$ = Vacuum permittivity = $8.854\times 10^{-12}\ \text{F/m}$

$A$ = Area = $10\times 2\times 10^{-4}\ \text{m}^2$

$d$ = Distance between plates = 1 mm

$V_c$ = Changed voltage = 60 V

$V$ = Initial voltage = 100 V

$R$ = Resistance = $1000\ \Omega$

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}$

We have the relation

$V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}$

The time taken for the potential difference to reach the required level is $1.62\times 10^{-8}\ \text{s}$ .

5 0

2 years ago