The question is incomplete, the complete question is;
A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?
Answer:
See explanation
Explanation:
From Newton's law of cooling;
θ1 - θ2/t = K(θ1 + θ2/2 - θo]
Where;
θ1 and θ2 are initial and final temperatures
θo is the temperature of the surroundings
K is the constant
t is the time taken
Hence;
100 - 60/5 = K(100 + 60/2 - θo)
100 - 40/10 = K(100 + 40/2 - θo)
8= (80 - θo)K -----(1)
6= (70 - θo)K -----(2)
Diving (1) by (2)
8/6 = (80 - θo)/(70 - θo)
8(70 - θo) = 6(80 - θo)
560 - 8θo = 480 - θo
560 - 480 = -θo + 8θo
80 = 7θo
θo = 11.4°
Again from Newton's law of cooling;
θ = θo + Ce^-kt
Where;
t= 0, θ = 60° and θo = 11.4°
60 = 11.4 + C e^-K(0)
60 - 11.4 = C
C= 48.6°
To obtain K
40 = 11.4 + 48.6e^-10k
40 -11.4 = 48.6e^-10k
28.6/48.6 = e^-10k
0.5585 = e^-10k
-10k = ln0.5585
k= ln0.5585/-10
K= 0.0583
Hence, the temperature in 15 minutes;
θ= 11.4 + 48.6e^(-0.0583 × 15)
θ= 31.7°
The cation that is reabsorbed from the urine in response to aldosterone
Answer:
If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.
A kind of variable that a researcher purposely changes in investigation is
Answer:
independent variable
Explanation:
A simple pendulum consists of a ball of mass 3 kg hanging from a uniform string of mass 0.05 kg and length L. If the period of oscillation of the pendulum is 2 s, determine the speed of a transverse wave in the string when the pendulum hangs vertically.
Answer:
v = 3.12 m/s
Explanation:
First, we will find the length of the string by using the formula of the time period:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\l = \frac{T^2g}{4\pi^2}\\\\[/tex]
where,
l = length of string = ?
T = time period = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]l = \frac{(2\ s)^2(9.81\ m/s^2)}{4\pi^2}\\\\l = 0.99\ m[/tex]
Now, we will find tension in the string in the vertical position through the weight of the ball:
T = W = mg = (3 kg)(9.81 m/s²)
T = 29.43 N
Now, the speed of the transverse wave is given as follows:
[tex]v=\sqrt{\frac{Tl}{m}}\\\\v=\sqrt{\frac{(29.43\ N)(0.99\ m)}{3\ kg}}\\\\[/tex]
v = 3.12 m/s
Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point
Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
Explain how blood circulation takes place in humans?
Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.
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A 500 kg rocket sled is coasting in reverse at 10 m/s (to the left). It then turns on its rocket engines for 10.0 s, with a thrust of 1500 N (to the right). What is its final velocity? (Remember velocity has magnitude and direction)
Explanation:
F = ma
[tex]a = \frac{f}{m} [/tex]
[tex]a = \frac{1500}{500} = 3[/tex]
[tex]a = \frac{v2 - v1}{t} [/tex]
[tex]3 = \frac{v2 - 10}{10} [/tex]
v2 (final) = 40 m/s to the right direction
If a negatively charged particle is placed inside a uniform electric field the electric force that will act on that particle points in what direction in reference to the electric field lines?
Answer:
opposite direction
Explanation:
An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.
Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.
Answer:
Explanation:
negatively charged particle is placed inside uniform electric field
The force on the charge due to the electric field is
F = q E
when the charge is negative so the force on the charge is opposite to the direction of electric field.
The electric field is opposite to the force.
the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop
if you jog at a speed of 1.5m/s for 20 seconds how far di you travel
Answer: 30m
Explanation:
Given:
Speed: 1.5m/s
Time: 20 seconds
Distance = speed × time
Distance = 1.5 × 20
= 30m
Therefore you will travel 30m
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Why we use semiconductor instead of metal in thermopile.
Answer:
Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm
The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %
The known values are;
The increase in pressure per 10 meter increase in depth = 1.0 atm
The depth of the deepest ocean = 12 km = 12,000 m
The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm
The unknown
The percentage the density of water increased in the deepest ocean
Strategy;
Find the pressure at the deepest point of the deepest ocean and apply the compressibility
We have;
[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]
The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm
Therefore, we have for one cubic meter of water
[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]
Therefore;
[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³
The new volume = V - [tex]\mathbf{\partial}[/tex]V
∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³
The initial density = mass/(1 m³)
The new density = mass/(0.96 m³)
The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;
[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]
∴ [tex]\mathbf{\partial}[/tex]ρ% = 4.1[tex]\mathbf {\overline 6}[/tex] %
The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %
Learn more about compressibility here;
https://brainly.com/question/18746977
Which factor affects kinetic energy but not potential energy?
Answer:
mass
Explanation:
because as the mass increase kinetic energy also increase
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
The coefficients of friction between a race cars tyres and the track surface are
the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).
Explanation:
please mark me as a brainlieast
b. Projectile on cliff (range)
An object of mass 5 kg is projected at an angle of 25° to the horizontal with a speed of 22 ms-1 from the top of the cliff.
The height of the cliff is 21 m. Take g, the acceleration due to gravity, to be 9.81 ms2
How far horizontally (to 1 decimal place) from the base of the cliff does the object land?
Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]
t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m
what is liquid pressure and its si unit?
The SI unit of pressure is the pascal: 1Pa=1N/m2 1 Pa = 1 N/m 2 . Pressure due to the weight of a liquid of constant density is given by p=ρgh p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.
Proper physical exercise makes bones _[blank 1]_.
People with stronger muscles and bones have better _[blank 2]_.
Which option shows the words that correctly fill in blank 1 and blank 2, in that order?
longer, flexibilitylonger, flexibility , ,
stronger, posturestronger, posture , ,
longer, posturelonger, posture , ,
stronger, flexibility
stronger, posturestronger, posture
hope that helped
physics approach to study macromoelcues at nanoscales
in detail plx
Answer:
Abstracto
Los ácidos nucleicos y las proteínas comprenden una red de biomacromoléculas que almacenan y transmiten información que sustenta la vida de la célula. El estudio de estos mecanismos es un campo llamado biología molecular. El desarrollo de esta ciencia siempre ha ido acompañado de avances técnicos que permiten romper barreras metodológicas para probar hipótesis novedosas. Entre los métodos disponibles para los biólogos moleculares, destacan cinco: electroforesis, secuenciación, clonación, transferencia y reacción en cadena de la polimerasa. Su impacto llega a la genética, la medicina y la biotecnología. Aquí, se revisan la relevancia histórica, los fundamentos técnicos y las tendencias actuales de estos cinco métodos esenciales. La revisión pretende ser útil tanto para estudiantes como para científicos profesionales que buscan adquirir conocimientos avanzados sobre el valor de estos métodos para investigar los mecanismos moleculares que sostienen la vida.
4. A diver is 20 m underwater and they are startled by a shark. They are tempted to take a big breath of air, drop their gear, and swim to the surface while holding their breath. Explain why this is dangerous g
Answer:
Explanation:
The air enters their lungs at the same pressure as the water at that depth.
If they hold their breath as they rise to atmospheric pressure, the expanding volume of air (due to decreasing pressure) trapped in their lungs will hyperextend the alveoli in their lungs, likely tearing blood lines and risking death by drowning in their own blood.
A capacitor consists of two parallel conducting plates, each of area 0.4 m2 and separated by a distance of 2.0 cm. Assume there is air between the plates. While connected to a battery the electric field within the plates is 500 N/C. The potential difference between the plates is: ________
a) 5.0 V
b) 10 V
c) 30 V
d) 20 V
Answer:
check photo
Explanation:
A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
Answer:
Resistance of the armature coil = 6.61 ohms
Back emf developed at normal speed = 98.62 V (Approx.)
Current drawn by the motor at one-third normal speed = 12.73 A
Explanation:
Given:
Potential difference V = 117 V
Current = 17.7 A
Motor drawn current = 2.78 A
Find:
Resistance of the armature coil
Back emf developed at normal speed
Current drawn by the motor at one-third normal speed
Computation:
A] Resistance of the armature coil R = V/ I
Resistance of the armature coil = 117 / 17.7
Resistance of the armature coil = 6.61 ohms
B] Back emf developed at normal speed = V- IR
Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)
Back emf developed at normal speed = 117 V - 18.37
Back emf developed at normal speed = 98.62 V (Approx.)
C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)
Current drawn by the motor at one-third normal speed = 17.7 - 4.97
Current drawn by the motor at one-third normal speed = 12.73 A
Why is it advised not to hold the thermometer by its bulb while reading it?
which of the following can not happen when a light ray strikes a new medium
Answer:
amplification
Explanation:
reflection can happen
some amount of lighr get absorbed
something gets refracted
but amplification cant
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot penetrate the membrane. The left side is filled with gas A and the right side is filled with gas B. We will assume that equipartition applies to both gases, but gas A has an excluded volume due to large molecules so its entropy has a different formula.
SA=NAkln(VA+ bNA)+f(UA,NA)
SB=NBkln(VB)+f(UB,NB)
Required:
If NA= 1 moles, NB = 2 moles, the total volume of the box is 1 m3, and b= 4 × 10-4 m3/mole, then find the equilibrium value of VA by maximizing the total entropy.
Answer:
The answer is "[tex]0.3336\ m^3[/tex]"
Explanation:
Using the Promideal gas law:
[tex]P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\[/tex]
[tex]\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\[/tex]
[tex]=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3[/tex]
The equilibrium value of Va is 0.3336 m³.
Ideal gas lawThe equilibrium value of Va is determine by applying ideal gas law as shown below;
Pressure of gas A = Pressure of gas B
Pa = Pb
Pa(Va - nab) = naRT----(1)
PbVb = nbRT -----(2)
Solve equation (1) and (2)
[tex]\frac{P_b}{RT} = \frac{n_b}{V_b} \\\\\frac{P_b}{P_a(V_a- n_ab)/n_a} = \frac{n_b}{V_b}\\\\\frac{n_a}{V_a - n_ab} = \frac{n_b}{V_b} \\\\\frac{V_a - n_ab}{V_b} = \frac{n_a}{n_b} \\\\\frac{V_a - b}{V_b} = \frac{1}{2}[/tex]
Va + Vb = 1
Vb = 1 - Va
[tex]\frac{V_a - b}{1 - V_a} = \frac{1}{2}[/tex]
2Va - 2b = 1 - Va
3Va = 1 + 2b
[tex]V_ a = \frac{1 + 2b}{3} \\\\V_a = \frac{1 + (2 \times 4\times 10^{-4})}{3} \\\\V_a = 0.3336 \ m^3[/tex]
Thus, the equilibrium value of Va is 0.3336 m³.
Learn more about equilibrium value here: https://brainly.com/question/22569960
why the stone moves away when the string is broken rotation
Answer:
When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path
Answer:
when the string's rotation is broken, there will be no centripetal force to keep the stone stationary. Thus, the stone will flung away when the rotation is stopped
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
The propeller on a boat motor is initially rotating at 8 revolutions per second. As the boat captain reduces the boat speed, the propeller SLOWS at a steady rate of 0.9 revolutions per second per second. After 17 revolutions, how fast is the propeller spinning in revolutions per second
Answer: [tex]5.77\ rps[/tex]
Explanation:
Given
Initial angular velocity is [tex]\omega_i=8\ rps[/tex]
rate of reduction [tex]\alpha=0.9 rev/s^2[/tex]
after 17 revolution i.e. [tex]\theta =17\ rev[/tex]
using [tex]\Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta[/tex]
Insert the values
[tex]\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps[/tex]
list at least types of motion
Answer:
These four are rotary, oscillating, linear and reciprocating. Each one moves in a slightly different way and each type of achieved using different mechanical means that help us understand linear motion and motion control.
(I got this off the web so credits to the rightful owner and I hope you have good day :)
