Answer and Explanation:
a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.
b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.
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
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
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.
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
Which factor affects kinetic energy but not potential energy?
Answer:
mass
Explanation:
because as the mass increase kinetic energy also increase
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
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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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:
Why is it advised not to hold the thermometer by its bulb while reading it?
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.
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
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]
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.
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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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.
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;
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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
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 :)
. 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:
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
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
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]
Can someone please help!!!
Answer:
W = F • ∆x
so for work to be done, a force and displacement has to be in the same direction. (Ex: a box is being pushed forward and it's also moving forward.)
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
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.
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:
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A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction between the sled and the incline is .45, What is the tension force between the block and the tree
Answer:
Explanation:
ASSUMING that block = sled AND that the rope is parallel to the slope.
The force acting parallel due to the weight is
13.6(9.81)sin35.5 = 77.475 N
The maximum friction force is
(0.45)13.6(9.81)cos35.5 = 48.877 N
If rope tension is T
77.475 - 48.877 < T < 77.475 + 48.877
28.6 N < T < 126 N
28.6 N will occur if the block is on the verge of sliding downhill
126 N will occur if the block is on the verge of sliding uphill
Could be any value between them.
A positron is a particle that has the same mass but opposite charge of an electron. An electron and a positron are shot directly toward each other by a particle accelerator. They start very far from each other, each with a kinetic energy of
Complete Question
The Complete Question is Attached below
Answer:
[tex]E_t=2.14*10^{-13}J[/tex]
Explanation:
From the question we are told that:
Kinetic energy [tex]K.E=5*10^{-14}[/tex]
Generally, the equation for Total Energy is mathematically given by
[tex]E_t=K.E+2me[/tex]
Where
Mass of an electron in MeV
[tex]m=0.510 998 950 00 MeV[/tex]
And
Electron Charge
[tex]1.6 * 10^{-19}[/tex]
Therefore
[tex]E_t=K.E+2me[/tex]
[tex]E_t=5*10^{-14}*0.510 998 950 00*1.6 * 10^{-19}[/tex]
[tex]E_t=2.14*10^{-13}J[/tex]
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 kind of variable that a researcher purposely changes in investigation is
Answer:
independent variable
Explanation:
