Answer:
B I believe
Explanation:
A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative relative to the outer surface. How much work is required to eject a positive sodium ion (Na ) from the interior of the cell?
Answer:
The workdone is [tex]W = 1.712 *10^{-20 } \ J[/tex]
Explanation:
From the question we are told that
The potential difference is [tex]V = 107 mV = 107 *10^{-3} \ V[/tex]
Generally the charge on [tex]Na^{+}[/tex] is [tex]Q_{Na^{+}} = 1.60 *10^{-19 } \ C[/tex]
Generally the workdone is mathematically represented as
[tex]W = Q_{Na^{+}}V[/tex]
=> [tex]W = 1.60 *10^{-19 } * 107 *10^{-3}[/tex]
=> [tex]W = 1.712 *10^{-20 } \ J[/tex]
As a rough model of the impact of walking/running, consider that half the mass of the body falls from a height of 4.77-cm onto a single foot. (During a typical stride, an adult's center-of-mass moves approximately this distance vertically). Use the kinematic equations to calculate the speed of an object falling from this height at the moment of impact with the ground under the influence of gravity.A. As a rough model of the impact of walking, consider that half of the mass of the entire body strikes the ground with a downward velocity of 1.0 m/s and comes to a full vertical stop over an impact duration of 20 ms. Calculate the force associated with this single step for a person with a mass of 74.2 kg. B. Calculate the stress (solid pressure) of a force of 1880 N applied across the 0.4 cm^2 cross-sectional area of the typical Achilles tendon. For reference, the maximum rupture stress of tendons has been reported in the range of 100-150 MPa.
Answer:
0.967 m/s
1855 N
[tex]46.375\ \text{MPa}[/tex]
Explanation:
v = Final velocity
u = Initial velocity = 0
s = Displacement = 4.77 cm
g = a = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the kinematic equations
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]
The velocity of the object at the moment of impact is 0.967 m/s
Now
[tex]\Delta v[/tex] = Change in velocity = 1 m/s
t = Time taken = 20 ms
m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]
[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]
The force associated with a single step of the person is 1855 N
A = Area = [tex]0.4\ \text{cm}^2[/tex]
Stress is given by
[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]
The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]
The speed of object during falling is 0.967 m/s.
(A) The magnitude of force associated with this single step for a person is 1855 N.
(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
Given data:
The height of fall is, h = 4.77 cm = 0.0477 m.
The magnitude of downward velocity is, v' = 1.0 m/s.
The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].
The mass of person is, m = 74.2 kg.
The magnitude of force is, F' = 1880 N.
The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].
The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,
[tex]v^{2}=u^{2}+2gh[/tex]
Solving as,
[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]
Thus, the speed of object during falling is 0.967 m/s.
(A)
Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.
Apply the expression of average force as,
[tex]F =\dfrac{m'v'}{t}[/tex]
Solving as,
[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]
Thus, the magnitude of force associated with this single step for a person is 1855 N.
(B)
The expression for the stress is given as,
[tex]\sigma = \dfrac{F'}{A}[/tex]
Solving as,
[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]
Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
Learn more about the Stress force here:
https://brainly.com/question/18274389
A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?
Answer:
(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰
(a)(ii) The maximum torque is 0.132 Nm
(b) The orientation of the coil is 45⁰
Explanation:
Given;
diameter of the circular wire, d = 8.6 cm = 0.086 m
radius of the wire, r = d /2 = 0.043 m
number of turns, N = 15 turns
magnetic field, B = 0.56 T
The torque on the wire is given by;
τ = NIABsinθ
where;
θ is the orientation of the wire
(a) maximum torque occurs when the orientation of the wire is at 90⁰
The maximum torque is given by;
τ = NIABsin(90⁰)
τ = NIAB
τ = (15)(2.7)(π x 0.043²)(0.56)
τ = 0.132 Nm
(b)
71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm
[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]
Suppose two parallel-plate capacitors have the same charge Q, but the area of capacitor 1 is A and the area of capacitor 2 is 2 A.If the spacing between the plates, d, is the same in both capacitors, and the voltage across capacitor 1 is V, what is the voltage across capacitor 2?Express your answer in terms of V but do not type in the symbol "V"
Answer:
V' = V/2
Explanation:
The voltage across a parallel plate capacitor is given as follows:
V = Q/C
where,
V = Voltage across capacitor
Q = Charge on Capacitor
C = Capacitance of Capacitor = A∈₀/d
Therefore,
V = Qd/A∈₀
where,
A = Area of plate
d = distance between plates
∈₀ = permittivity of free space
FOR CAPACITOR 1:
Q = Q
d = d
A = A
V = V
Therefore,
V = Qd/A∈₀ --------------- equation (1)
FOR CAPACITOR 2:
V' = ?
Q' = Q
d' = d
A' = 2A
Therefore,
V' = Q'd'/A'∈₀
V' = Qd/2A∈₀
V' = (1/2)(Qd/A∈₀)
using equation (1):
V' = V/2
Two identical bars are conducting heat from a region of higher temperature to one of lower temperature. In arrangement A, the bars conduct 80 J of heat in a certain amount of time. How much heat is conducted in B during the same time
Answer:
Q' = 320 J
Explanation:
The arrangements are given in attachment. The Fourier's Law of Heat Conduction states that:
Q = KAΔT/L
where,
Q = Heat Transferred
K = Constant (Conduction Coefficient)
A = Surface Area of Heat Transfer
ΔT = Difference of Temperature between two surfaces
L = Length between surfaces
For Arrangement AL
Q = 80 J
Therefore,
80 = KAΔT/L ------------- equation (1)
Now, for arrangement B:
A' = 2 A (As, the rods are now connected in parallel with each other)
L' = L/2
Therefore,
Q' = K(2A)ΔT/(L/2)
Q' = 4 KAΔT/L
using equation (1)
Q' = 4(80 J)
Q' = 320 J
A rod is pivoted about its center and oriented horizontally. A 5.0-N force directed upward is applied 4.0 m to the left of the pivot and another upward 5.0-N force is applied 1.5 m to the right of the pivot. What is magnitude of the total torque about the pivot?
Answer:
The total torque is 27.5 Nm
Explanation:
Given;
5.0-N force directed upward is applied 4.0 m to the left of the pivot,
5.0-N force directed upward is applied 1.5 m to the right of the pivot,
Taking the moment about the pivot, the total torque is given by;
τ = Fr
where;
F is the appllied force
r is the radius of the force arm
τ = (5 N x 4 m) + (5 N x 1.5 m)
τ = 27.5 Nm
Therefore, the total torque is 27.5 Nm
WHAT IS TRANS ATLANTIC SLAVE TRADE
Suppose a certain object has a mass of 5.00 kilograms on the earth. On the
Moon, where g is 1.6 m/s/s what would its mass be?*
Answer:
it would be 49.03325 Newton.
What
A moving object always has energy in its
A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.3 m/s. The drag force is of the form bv2 What is the value of b?
Answer:
The value is [tex]b = 0.00026 \ kg / m[/tex]
Explanation:
From the question we are told that
The mass of the Ping-Pong is [tex]m = 2.3 \ g = 0.0023 \ kg[/tex]
The terminal speed is [tex]v = 9.3 \ m/s[/tex]
The drag force is [tex]bv^2[/tex]
Generally the resultant force on the Ping- Pong is mathematically represented as
[tex]F = mg - bv^2[/tex]
when terminal velocity is attained , the resultant force is zero so
[tex]0 = mg - bv^2[/tex]
=> [tex]b = \frac{m * g}{v^2}[/tex]
=> [tex]b = \frac{0.0023 * 9.8}{ 9.3 ^2}[/tex]
=> [tex]b = 0.00026 \ kg / m[/tex]
Which image illustrates the interaction of a light wave with a mirror?
t J
A
с
.
A. A
B. B
C. C
D. D
0
Answer:
I'm pretty sure its A
Explanation:
because its a reflection- Hope you get a good grade!
How do lenses and mirrors compare in their interactions with light?
A. Lenses spread apart light; mirrors do not.
B. Lenses reflect light; mirrors do not.
C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.
Answer:
C. lenses refract light; mirrors do not
This question involves the concepts of reflection and refraction.
The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".
LENSES AND MIRRORSWhen it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the bending of light rays, while passing through a medium, without any rebound or absorption.
Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.
Learn more about reflection and refraction here:
https://brainly.com/question/3764651
You discover a binary star system in which one member is a15MSun main-sequence star and the other star is a 10MSun giant. How do we believe that a star system such as this might have come to exist?
Answer:
Explanation:
The giant star must have at least once been the more massive star and then subsequently transferred some of its mass to its companion, the other star.
The two stars would be around the same age, so the more massive one would have turned into a giant first before the other one did or even had a chance to
car driving on a circular test track shows a constant speedometer reading of 100 kph for one lap. a. Describe the car's speed during this time. b.
Answer:
Speed = 100 km/h
Explanation:
Given:
Speedometer reading = 100 kph for one lap
Assume;
Time taken to complete one lap = 1 hour
Computation:
Speed = Distance / Time
Speed = 100 / 1
Speed = 100 km/h
A hydraulic car jack needs to be designed so it can lift a 2903.57 lb car assuming that a person can exert a force of 24.41 lbs. If the piston the person is pushing on had a radius of 3.26 cm, what should the diameter of the piston be that is used to raise the car?
Answer:
Diameter of the piston would be 0.71 m (71.1 cm)
Explanation:
From the principle of pressure;
[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
Let [tex]F_{1}[/tex] = 2903.57 lb, [tex]F_{2}[/tex] = 24.41 lbs, [tex]r_{2}[/tex] = 3.26 cm = 0.0326 m.
[tex]A_{2}[/tex] = [tex]\pi r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] x [tex](0.0326)^{2}[/tex]
= 0.00334 [tex]m^{2}[/tex]
So that:
[tex]\frac{2903.57}{A_{1} }[/tex] = [tex]\frac{24.41}{0.00334}[/tex]
[tex]A_{1}[/tex] = [tex]\frac{2903.57*0.00334}{24.41}[/tex]
= 0.3973
[tex]A_{1}[/tex] = 0.4 [tex]m^{2}[/tex]
The radius of the piston can be determined by:
[tex]A_{1}[/tex] = [tex]\pi r^{2}[/tex]
0.3973 = [tex]\frac{22}{7}[/tex] x [tex]r^{2}[/tex]
[tex]r^{2}[/tex] = [tex]\frac{0.3973*7}{22}[/tex]
= 0.1264
r = [tex]\sqrt{0.1264}[/tex]
= 0.3555
r = 0.36 m
Diameter of the piston = 2 x r
= 2 x 0.3555
= 0.711
Diameter of the piston would be 0.71 m (71.1 cm).
Section 4.1- Newton's First Law
Answer:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. ... If that velocity is zero, then the object remains at rest.
Explanation:
Answer:
Newton's First Law is about inertia; objects at rest stay at rest unless acted upon and objects in motion continue that motion in a straight line unless acted upon. The amount of inertia an object has is simply related to the mass of the object.
What becomes V if we use 2 resistors of 4W in parallel?
A. 2.66 V
B. 6 V
C. 12 V
D. 24 V
Answer:
This question is incomplete.
Explanation:
This question is incomplete. However, it should be noted that the voltage, V, across resistors in parallel is the same (although there currents are not the same). Thus, if a voltage has been provided, it remains the same but if not provided, you can solve for it using the formulas below
V = IR
where V is the voltage. I is the current and R is the resistance
R in parallel can be calculated as R = 1/R₁ + 1/R₂ + 1/R₃ + ......
A crate of books rests on a level floor. To move it along the floor at a constant velocity, why do you exert less force if you pull it at an angle Ï above the horizontal than if you push it at the same angle below the horizontal?
Answer:should be a matter of vector analysis.
Pulling above the horizontal has less surface area for the opposing friction
Explanation:
A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2370 Pa greater than atmospheric pressure? The density of glycerine is 1.26X10^3 kg/m^3
Answer:
So, at the depth of 24 cm below the surface of the glycerine the pressure is 2970 Pa. Hence, this is the required solution.
Explanation:
Given that,
Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.
The density of glycerine,
We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :
h = 0.24 meters
or
h = 24 cm
what is primary purpose of Pathfit?
Answer:
to show the arts and creativity of the person and to show also the culture of the place..
Explanation:
An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 and the temperature remains constant?
Answer:
2.22 kPaExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
From the question we have
[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]
We have the final answer as
2.22 kPaHope this helps you
In a certain time, light travels 3.50 km in a vacuum. During the same time, light travels only 2.35 km in a liquid. What is the refractive index of the liquid?
Answer:
1.45
Explanation:
Refractive index of the liquid is given as;
Refractive index = [tex]\frac{speed of light in vacuum}{speed of light in liquid}[/tex]
But,
speed = [tex]\frac{distance}{time}[/tex]
Since a certain light of specific wavelength was used during the same time, let the time be represented by t.
So that;
speed of light in vacuum = [tex]\frac{3500}{t}[/tex]
speed of light in the liquid = [tex]\frac{2350}{t}[/tex]
Refractive index = [tex]\frac{3500}{t}[/tex] ÷ [tex]\frac{2350}{t}[/tex]
= [tex]\frac{3500}{t}[/tex] x [tex]\frac{t}{2350}[/tex]
Refractive index = [tex]\frac{3500}{2350}[/tex]
= 1.4536
= 1.45
The refractive index of the liquid is 1.45.
why is it more painful to walk on gravel with your shoes off then on (3 marks please)
Answer:
Because shoes protect our feet from some of the most harmful platforms
Gravel has some small pebbles on it sometimes (or other sharp objects)
Gravel is pretty hard.
?
Which activity is health enhancing?
folding a load of laundry
driving long distances
O biking to school
unloading the dishwasher
Answer:
biking to school
Explanation:
plato
In the Bohr model of the hydrogen atom, an electron in the 1st excited state moves at a speed of 2.19 106 m/s in a circular path having a radius of 5.29 10-11 m. What is the effective current associated with this orbiting electron?
Answer:
I = 1.05x10⁻³ A
Explanation:
By definition, an electric current is the rate of charge flow at a given time:
[tex] I = \frac{q}{t} [/tex]
Where:
q: is the electrons charge = 1.602x10⁻¹⁹ C
t: is the time
In a circular motion, the time is given by:
[tex] t = T = \frac{2\pi}{\omega} = \frac{2\pi}{v/r} = \frac{2\pi r}{v} [/tex]
Where:
ω: is the angular speed = v/r
v: is the speed = 2.19x10⁶ m/s
r: is the radius = 5.29x10⁻¹¹ m
[tex] t = \frac{2\pi r}{v} = \frac{2\pi 5.29 \cdot 10^{-11} m}{2.19 \cdot 10^{6} m/s} = 1.52 \cdot 10^{-16} s [/tex]
Now, the effective current is:
[tex] I = \frac{q}{t} = \frac{1.602 \cdot 10^{-19} C}{ 1.52 \cdot 10^{-16} s} = 1.05 \cdot 10^{-3} A [/tex]
Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.
I hope it helps you!
A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.173. (indicate the direction with signs of your answer) (a) What is the force of kinetic friction in N acting on the block? (b) What is the block's acceleration in /s^2? (c) How Far will it slide (in m) before coming to rest? Plz answer as soon as possible
Answer:
Explanation:
a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force
R = mg where m is mass of the block
Force of friction F = μ x mg
= .173 x 12.2 x 9.8
= 20.68 N
b ) Only force of friction is acting on the body so
deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²
acceleration = - 1.7 m /s²
c )
v² = u² - 2 a s
v = 0 , u = 3.9 m /s
a = 1.7 m /s
0 = 3.9² - 2 x 1.7 x s
s = 4.47 m
a book weighing 1.0 newton is lifted 2m. how much work was done?
Answer:
Work done, W = 2 J
Explanation:
Given that,
Weight of a book, W = F = 1 N
It is lifted to a height of 2 m
We need to find the work done. It can be calculated using the formula as follows :
W = F d
Put all the values,
W = 1 N × 2 m
W = 2 J
So, 2J of work was done.
On the image at right, the two magnets are the same. Which paper clip would be harder to remove?
Answer:B
Explanation: The book is thinner making magnets attraction stronger, making the paper clip harder to move
You are driving on the highway at a speed of 40 m/s (which is over the speed limit) when you notice a cop in front of you. To avoid a ticket, you press on the brake and slow to a speed of 30 m/s over the course of 5 seconds. What is the acceleration of the car? WORK=BRAINLIEST
What is your car's initial velocity?
What is your car's final velocity?
How long does it take the car to slow down?
Write the equation you will use to solve this problem.
What is the acceleration of your vehicle?
+ 2.0 m/s^2
- 2.0 m/s^2
+ 8.0 m/s^2
- 6.0 m/s^2
Explanation:
U = 40m/s
V = 30m/s
T = 5 sec
A = ?
[tex]a = \frac{u - v}{t}[/tex]
[tex]a = \frac{40 - 30}{5}[/tex]
[tex]a = \frac{10}{5}[/tex]
[tex]a = 2[/tex]
since it's decreasing in speed, The acceleration will be " - 2.0ms^-2 " or " - 2.0m/s^2 "
If you liked this answer, feel free to follow me for more!
Btw don't mind me answering twice. I want the free points and maybe another brainliest? lol.
Which one would be felt louder: a sound with a SIL of 60dB and a frequency of 40Hz or a sound with a SIL of 60dB and a frequency of 100Hz ? a) they both will appear to have the same loudness b) sound with a SIL of 60dB c) sound with a SIL of 60dB and a frequency of 1000Hz d) whichever is sounded first e) cannot say and a frequency of 40ã
Answer:
a) they both will appear to have the same loudness
Explanation:
When we talk about how loud a sound is , we describe it in decibels (dB). The value of this unit on measurement in this question is what we are going to build our answer on.
Their sound intensity level is the same at 60db even though each of these separate sounds have different frequencies of 40Hz and 100Hz respectively.
The both of them will therefore have the same loudness actually as their sound intensity level has no difference bat 60db each.