1. Estimate the buoyant force that air exerts on a man. (To do this, you can estimate his volume by knowing his weight and by assuming that his weight density is about equal to that of water. Assume his weight is 940 N.) answer in N
2.On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 17000 N.
(a) What is the weight of the displaced air?
answer in N
(b) What is the volume of the displaced air?
answer in m^3

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0  = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.  

So, option B is the answer

Answer:

Explanation:

The formula for angular velocity is

[tex]\omega=\frac{\theta}{t}[/tex] where omega is the angular velocity, theta is the change in the angular rotation, and t is the time in seconds. First and foremost, we have the angular rotation in minutes and the time in seconds, so that's a problem we have to amend. Let's change the angular rotation to rotations per second:

[tex]281.133\frac{r}{min}*\frac{1min}{60s}=4.68555\frac{r}{s}[/tex]

Now we're ready to set up the problem:

[tex]4.68555=\frac{\theta}{3.686}[/tex] and we multiply both sides by 3.686 to get the rotations per seconds:

θ = 17.27 rotations

Answer:

Octabromine tetraphosphide

Explanation:

This compound has in its formula:

- Eight bromines

- Four phosphorous

8 → octa prefix

4 → tetra prefix

Right answer is Octabromine tetraphosphide

science is all about the world around us

Answer:

c) charges exert forces on other charges.

Explanation:

When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.

Explanation:

For air, n1 = 1.00003; for water, n2 = 1.3330

Given: θ2 = 30 degrees, then

θ1 = arcsin [(n2/n1) sin θ2]

= arcsin [(1.3330/1.0003) sin (40)]

= 58.93 degrees

Note that since, in this example, light is traveling from a medium of higher density (water; n2 = 1.3330) to a medium of lower density (air; n1 = 1.0003), then n2 > n1, and the angle of refraction (θ1) is larger than the angle of incidence (θ2), thus the light bends away from the normal (in this example, the vertical) as it leaves the water and enters the air.

Answer:

The correct answer is "[tex](0,300\times 10^{-3} \ N/C)[/tex]".

Explanation:

The given problem seems to be incomplete. Please find the attachment of the complete query.

According to the question,

At point A, we have

⇒ [tex]E_x = \frac{k q_1}{d_1^2} Cos \theta_1 - \frac{k q_2}{d_2^2} Cos \theta_2[/tex]

or,

⇒ [tex]E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ][/tex]

         [tex]=0[/tex]

and,

⇒ [tex]E_y = \frac{kq_1}{d_1^2}Sin \theta_1 +\frac{kq_2}{d_2^2}Sin \theta_2[/tex]

or,

⇒ [tex]E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ][/tex]

         [tex]=0.3 \ N/C[/tex]

Answer:

R = 0.0015Ω

Explanation:

The formula for calculating the resistivity of a material is expressed as;

ρ = RA/l

R is the resistance

ρ is the resistivity

A is the area of the wire

l is the length of the wire

Given

l = 85cm = 0.85m

A = πr²

A = 3.14*0.0018²

A = 0.0000101736m²

ρ = 1.75 × 10-8Ωm.

Substitute into the formula

1.75 × 10-8 = 0.0000101736R/0.85

1.4875× 10-8 = 0.0000101736R

R = 1.4875× 10-8/0.0000101736

R = 0.0015Ω

Answer:

the answer should be b) 3.8 x 10-5 v

Answer:

the tendency of electron dipole moments to align with an applied magnetic field

Explanation:

Paramagnetism has to do with possession of unpaired electrons. Substances that possess unpaired electrons are said to be paramagnetic.

When an external magnetic field is applied to a paramagnetic substance, the magnetic field causes the electrons spins of the paramagnetic substance to align parallel to the field,which leads to a net attraction.

Hence, paramagnetism is closely associated with the tendency of electron dipole moments to align with an applied magnetic field.

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω

Answer:

Three Types of Computer The Computer are classified into three main types:: • Analog Computers • Digital Computers • Hybrid Computers (Analog + Digital) Analog Computers:: Analog Computer measures “Physical Quantities” for example Temperature, Voltage, Pressure, and Electric Current.

Answer:

40 because if it is the same weight then there is no weight to make the ride slower so it 40

Explanation:

Solution :

Let [tex]$d_1=\frac{5.5}{2}[/tex]

          = 2.75 m

[tex]d_2 = 0.21 \ m[/tex]

And [tex]$d=|d_1-d_2|$[/tex]

       [tex]$d=(d_1+d_2) - (d_1-d_2)$[/tex]

       [tex]$d=(2.75+0.21) - (2.75-0.21)$[/tex]

       [tex]$d = 2.96-2.54$[/tex]

       [tex]d = 0.42 \ m[/tex]

a). At minimum,

[tex]$d=\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2d$[/tex]

  = 2 x 0.42

  = 0.84 m

Frequency, [tex]$\nu = \frac{v}{\lambda}$[/tex]

                      [tex]$=\frac{340}{0.84}$[/tex]

                      = 404.76 Hz

Therefore, the frequency of he sound, [tex]$\nu$[/tex] = 404.76 Hz

b). At maximum, λ = d = 0.42 m

Therefore, the frequency, [tex]$\nu = \frac{v}{\lambda}[/tex]

                                             [tex]$=\frac{350}{0.42}$[/tex]

                                             = 809.52 Hz

Answer:

Explanation:

That is an amazing fact.

The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.

The answer is D

Answer:

Standard units are commonly used units of measurement, which help us measure length, height, weight, temperature, mass and more. These units are standardised, which means that everyone gets the same understanding of the size, weight and other properties of objects and things.

Explanation:

Answer:

233.33°F

Explanation:

(385K - 273.15) * 9/5 + 32 = 233.33°F

Answer:

 emf = 312 V

Explanation:

In this exercise the electromotive force is asked, for which we must use Faraday's law

           emf =  [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt

           Ф = B. A = B A cos θ

bold type indicates vectors.

They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values ​​1

It also indicates that the area is reduced from  a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear

            emf = -N B [tex]\frac{dA}{dT}[/tex]

            emf = - N B (A_f - A₀) / Dt

we calculate

           emf = - 60 1.60 (0 - 0.325) /0.100

           emf = 312 V

The direction of this voltage is exiting the page

Answer:

a)  W = - 6.825 J,  b) θ = 1.72 revolution

Explanation:

a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

         W = ΔK

         W = K_f - K₀

          W = ½ m v_f² - ½ m v₀²

         W = ½ 0.325 (5.5² - 8.5²)

         W = - 6.825 J

b) find us the coefficient of friction

Let's use Newton's second law

            fr = μ N

y-axis (vertical)   N-W = 0

            fr = μ W

work is defined by

             W = F d

the distance traveled in a revolution is

             d₀ = 2π r

             W = μ mg d₀ = -6.825

            μ = [tex]\frac{ -6.825}{d_o \ mg}[/tex]

The total work as the object stops the final velocity is zero v_f = 0

         W = 0 - ½ m v₀²

          W = - ½ 0.325 8.5²

          W = - 11.74 J

           μ mg d = -11.74

we subtitle the friction coefficient value

           ( [tex]\frac{-6.8525 }{d_o mg}[/tex]) m g d = -11.74

               6.825  [tex]\frac{d}{d_o}[/tex] = 11.74

               d = 11.74/6.825  d₀

               d = 1.7201  2π 0.400

               d = 4.32 m

this is the total distance traveled, the distance and the angle are related

              θ = d / r

              θ = 4.32 / 0.40

              θ = 10.808 rad

we reduce to revolutions

              θ = 10.808 rad (1rev / 2π rad)

              θ = 1.72 revolution

This question deals with projectile motion, which is a motion on both the x-axis and y-axis, simultaneously. The total time of flight of the projectile trajectory is given, while the time to reach the highest point of the projectile is required to be found.

The projectile will reach the highest point in "3 seconds".

The total time of flight of a projectile is the time during which the projectile remains in the air. For a projectile motion that ends up on the same horizontal level, from where it started, the time to reach the highest point, is equal to half of the total time of flight.

In other words, the projectile motion takes the same time, to go from the starting level to the highest point (i.e upward motion), as the time taken to reach the starting level from the highest point (i.e downward motion).

[tex]t = \frac{1}{2}T[/tex]

where,

t = time to reach the highest point = ?

T = total time of flight = 6 seconds

Therefore,

[tex]t - \frac{1}{2}(6\ seconds)[/tex]

t = 3 seconds

Learn more about the projectile motion here:

https://brainly.com/question/20689870?referrer=searchResults

A  5

v=  dt/ dx  =−5+12t

Initial velocity means at t=0, which is −5+0=−5.

Thus, −v=5n

Answer:

approaches infinity

Explanation:

There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.

The relativistic momentum can be written as:

[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

where

u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.

m = mass of the object

c = speed of light.

So, as u tends to c, we will have:

[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.

Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.

Answer:

independent variable

Explanation:

Answer:

Problem solving

hope this helps :)

Answer:

C . Magnetic dipoles is the correct

Answer:

b). movement of charged particles.

Explanation:

These charges create the nagnetic dipoles.

Answer:

1. The image is real

2. 5.85

3. h' = 3.05 mm

4. The image is upright

Explanation:

1. Start with the first lens and apply 1/f = 1/p + 1/q

1/5.01 = 1/13.7 + 1/q

q = 7.90 cm

Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,

1/25.9 = 1/54.6 + 1/q

q = 49.3 cm behind the second lens

Using that information, since q is positive, the image is real

2. Also, using that information, you have the second answer, which is 49.3 cm

The height can be found from the two magnifications.

m = -q/p

m1 = -7.9/13.7 = -.577

m2 = -49.3/54.6 = -.903

Net m = (-.577)(-.903) = .521

Then, m = h'/h

.521 = h'/5.85

3. h' = 3.05 mm

4. For the fourth answer, since the overall magnification is positive, the final image is upright

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]

Therefore,

[tex]$AD-BD = \frac{\lambda}{2}[/tex]

[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2 \times 0.4985$[/tex]

[tex]$\lambda = 0.99714 \ m$[/tex]

Thus the frequency is :

[tex]$f=\frac{v}{\lambda}$[/tex]

[tex]$f=\frac{340}{0.99714}$[/tex]

[tex]f=340.9744[/tex] Hz