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Ksenya-84 [330]
3 years ago
12

Answer please 98 points also brainliest

Physics
2 answers:
notsponge [240]3 years ago
6 0

Answer:

d.

Explanation:

Look at it this way, when a sponge is saturated it cannot take in any more water. When air (a mixture of gasses) is saturated it cannot take in any more water vapour and water vapour will condense out as liquid water. This is termed the saturation point.

Air is classed usually as a percentage of saturation, dry air being 0% and saturated 100%

However, unlike the sponge ananalogy, 1kg of air will have a different capacity for water vapour depending on its temperature (and pressure). Charts (Psycrometric Charts) are available that allow the degree of saturation to be calculated, usually by determining the dew point (temperature at which 100% saturation occurs) by using a wet bulb thermometer and comparing it with the normal dry bulb temperature.

Yuri [45]3 years ago
5 0

the answer is D) The air in the aquarium is saturated

none of the other answers were realistic AND I took this before so I know the right answer.

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A small toy cart equipped with a spring bumper rolls toward a wall with a speed of v . The cart rebounds from the wall, with the
anastassius [24]

Using the initial momentum vector as a basis, the change in momentum vector Δp for the cart is drawn as shown in the attachment.

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial speed of cart = v_i = v

Final speed of cart = v_f = v

<u>Unknown:</u>

The change in momentum of cart = I  = ?

<u>Solution:</u>

I = \Delta p

I = p_f - p_i

I = mv_f - mv_i

I = m ( v_f - v_i )

I = m ( -v - v )

I = m ( -2v )

I = -2mv

I = -2p_i

\texttt{ }

<em>From the results above, we can conclude that the change in momentum vector Δp is twice the initial momentum vector p_i but in opposite direction.</em>

The vector <em>Δp could be drawn as shown </em><em>in the attachment.</em>

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

#LearnWithBrainly

5 0
3 years ago
Two solid spheres of radius R made of the same type of steel are placed in contact. The magnitude of the gravitational force the
dimulka [17.4K]
F1 = G * m*m / [2r]  ^2 = G * m^2 / 4r^2

Given the the mass is proportional to the volume and the volume is proportional to r^3, the mass of each big sphere is  3^3 = 27 times the mass of the small spheres. And the distance between the centers is 3r + 3r = 6r

F2 = G * [27m * 27m]/* [6r]^2 = G* 27^2 m^2 / 36r^2 = 81 G*m^2 / 4r^2 = 81*F1

Therefore, F2 = 81F1

Answer: option D) 81F1.
3 0
3 years ago
A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t
Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

\omega= 19 rev/s

we need to convert it into radiands per second. We know that

1 rev = 2 \pi rad

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

\theta= \omega t

where

\omega=119.3 rad

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

\theta=(119.3 rad/s)(5 s)=596.5 rad

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

\alpha = 1.6 rad/s^2

So the final angular speed will be given by

\omega_f = \omega_i + \alpha \Delta t

where

\omega_i = 119.3 rad/s is the initial angular speed

\alpha = 1.6 rad/s^2 is the angular acceleration

\Delta t = 15 s - 10 s = 5 s is the time interval

Solving the equation,

\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2

Substituting the numbers into the equation, we find

\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad

(e) 222 m

The instantaneous speed of the center of the wheel is given by

v_{CM} = \omega R (1)

where

\omega is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s

And so the displacement of the center of the wheel will be

d=v_{CM} t = (44.4 m/s)(5 s)=222 m

8 0
3 years ago
What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of t
yawa3891 [41]

Given Information:

Diameter of spherical cell = 0.040 mm

thickness = L = 9 nm

Resistivity =  ρ = 3.6×10⁷ Ω⋅m

Dielectric constant = k = 9.0

Required Information:

time constant = τ = ?

Answer:

time constant = 2.87×10⁻³ seconds

Explanation:

The time constant is given by

τ = RC

Where R is the resistance and C is the capacitance.

We know that resistivity of of any material is given by

ρ = RA/L

R =  ρL/A

Where area of spherical cell is given by

A = 4πr²

A = 4π(d/2)²

A = 4π(0.040×10⁻³/2)²

A = 5.026×10⁻⁹ m²

The resistance becomes

R =  (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹

R = 6.45×10⁷ Ω

The capacitance of the cell membrane is given by

C = kεoA/L

Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m

C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹

C = 44.5 pF

C = 44.5×10⁻¹² F

Therefore, the time constant is

τ = RC

τ = 6.45×10⁷*44.5×10⁻¹²

τ = 2.87×10⁻³ seconds

6 0
3 years ago
A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The
Temka [501]

Answer:

1m/s^2

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

ma=F-f

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

10a=20-10=10

a=\frac{10}{10}=1 m/s^2

Hence, the acceleration of the block=1m/s^2

4 0
3 years ago
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