Antibodies are parts of a wave that do not move when the wave is in phase? T OR F

An object weighs 15N in air and 13N when completely

anastassius [24]

Answer:

The volume of water displaced = 0.204 m³

Explanation:

From Archimedes' principle,

Upthrust = lost in weight = weight of water displaced.

U = W₁ - W₂..................... Equation 1

Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.

Given: W₁ = 15 N, W₂ = 13 N

Substituting these values into equation 1

U = 15 - 13

U = 2 N

But,

Weight of water displaced = mass of water displaced. × acceleration due to gravity.

Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

8 0

3 years ago

We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength

san4es73 [151]

Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

$\theta=1.22\frac{\lambda}{b}$

θ = 50.0*10^-7 rad

λ: wavelength of the light = 550nm

b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

$b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm$

hence, the smallest diameter objective lens is 13.4cm

8 0

3 years ago

Read 2 more answers

Can someone help me?!!!!!

german

<h2>Hello!</h2>

The answer is:

The first option, the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

$Distance=NorthCoveredDistance+EastCoveredDistance$

$Distance=4*180m+3*180m=720m+540m=1260m$

Actual distance between the start and the end point (displacement):

$ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m$

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

$DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m$

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0

3 years ago

During which two time intervals does the particle undergo equal displacement?

san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, $d_1=\dfrac{1}{2}\times 2\times 10=10\ m$

Area under BC, $d_2=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under CD, $d_3=\dfrac{1}{2}\times 2\times 7=7\ m$

Area under DE, $d_4=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under EF, $d_5=\dfrac{1}{2}\times 2\times 3=3\ m$

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

4 0

3 years ago

A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v

Tpy6a [65]

Current will be $I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz$
now just pluf in the values and Voila..

7 0

3 years ago