Answer:
f = 96.56 Hz
Explanation:
A soda bottle can be simulated by a system of a tube with one end open and the other closed. In this case at the closed end we have a node and at the open end a maximum, so the wavelength is
λ = 4L / n
where n = 1, 3, 5, ...
for our case the speed of sound is
v = λ f
we substitute
f / v = 4L / n
calculate
290 / (340 4) = L / n
0.213 = L / n
In the exercise the value of n is not indicated, but we can assume that it is the fundamental answer, therefore n = 1
L = 0.213 m
now we can calculate the frequency of the following resonance n = 3
f = 4L v / n
f = 4 0.213 340/3
f = 96.56 Hz
When a star is moving away from Earth, does the color of the star actually change? Explain - PLZZ HELP:)
Answer:
Yes the color changes. If the star is moving away from the earth the color of the star is shifted towards red (red-shift).
This is similar to the change in sound from an approaching train whistle (higher pitch) or a receding train whistle (lower pitch) - the Doppler Effect.
Note that the equations for the Doppler Effect for sound and for light are different - sound travels thru air but no such medium can be identified for light.
The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth is 1.5multiply1011 m. (a) Calculate the magnitude of the gravitational force exerted by the Sun on the Earth. N (b) Calculate the magnitude of the gravitational force exerted by the Earth on the Sun.
Answer:
a) 3.56 x 10^22 N
b) 3.56 x 10^22 N
Explanation:
Mass of the sun M = 2 x 10^30 kg
mass of the Earth m = 6 x 10^24 kg
Distance between the sun and the Earth R = 1.5 x 10^11 m
From Newton's law,
F = [tex]\frac{GMm}{R^2}[/tex]
where F is the gravitational force between the sun and the Earth
G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
m is the mass of the Earth
M is the mass of the sun
R is the distance between the sun and the Earth.
Substituting values, we have
F = [tex]\frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2}[/tex] = 3.56 x 10^22 N
A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N
b) The force exerted by the Earth on the Sun = 3.56 x 10^22 N
Part A:
The magnitude of the gravitational force exerted by the Sun on the Earth is :
Mass of the sun M = 2 x 10^30 kgMass of the Earth m = 6 x 10^24 kgDistance between the sun and the Earth R = 1.5 x 10^11 mFrom Newton's law,
F = GmM/F
G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2m is the mass of the EarthM is the mass of the sunR is the distance between the sun and the Earth.Substituting values,
F = = 3.56 x 10^22 N
The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N.
Part B:
The force exerted by the Earth on the Sun = 3.56 x 10^22 N.
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It is often illustrated in art, popular culture and everyday conversation that the Sun is Yellow. a. Using Wien’s Law and what you learned in class about the Sun’s temperature, explain why this is not true in reality. (20 points) b. In reality, what is the Sun’s true color? Why is this true? (20 points)
Answer:
a) surface λ = 5 10⁻⁷ m
nuecleus λ= 2,890 10⁻⁹ m
there is a mixture of wavelengths giving rise to white light
b) True color sun is White
Explanation:
Wien's law establishes a relationship between the temperature of a star and the maximum wavelength emitted
λ T = 2,898 10⁻³
if we calculate the wavelength of the Sun
T = 5800K λ = 5 10⁻⁷ m = 5000 nm
this is the surface temperature
in the part of the internal Sunspart (Nucleus) about 10⁶ degrees this is where the nuclear reaction occurs
in this region the photons are very short λ
λ = 2,890 10⁻³ / 106
λ= 2,890 10⁻⁹ m
these photons are absorbed and emitted on their way to the surface of the sun many times, therefore there is a mixture of wavelengths giving rise to white light
b) Sunlight when it reaches Earth is absorbed by atmospheric gases, mainly Blue, as it is absorbed by Nitrogen, it lets through mainly red and red
True color sun is White
A cube at 333 K contains two metals: 8.00 kg of solid Silver and 15.0 kg of solid Gold. It is placed in contact with a block of solid Iron at 1737 K. The system reaches equilibrium at 1337 K (all the silver and all the gold has melted) Find the mass of the iron. (cgold-liquid= 0.150 kJ/(kg K), csilver-liquid= 0.280 kJ/(kg K)
Answer:
Mass of Iron is 24.45 kg
Explanation:
Given that:
Mass of Silver, [tex]m_{S}[/tex] = 8.00 kg
Mass of Gold, [tex]m_{G}[/tex] = 15.0 kg
Initial temperature of Silver and Gold = 333 K
Initial temperature of Iron = 1737 K
Final temperature = 1337 K
Specific heat capacity of Gold-liquid, [tex]c_{G}[/tex] = 0.150 kJ/(kg K)
Specific heat of Silver-liquid, [tex]c_{S}[/tex] = 0.280 kJ/(kg K)
Known: Specific heat capacity of Iron, [tex]c_{I}[/tex] = 0.461 kJ/(kg K)
Therefore;
Heat lost by Iron = Heat gained by Silver + Heat gained by Gold
[tex]m_{I}[/tex] x [tex]c_{I}[/tex] x Δθ = ([tex]m_{S}[/tex] x [tex]c_{S}[/tex] + [tex]m_{G}[/tex] x [tex]c_{G}[/tex]) x Δθ
[tex]m_{I}[/tex] x 0.461 x (1737 - 1337) = (8 x 0.280 + 15 x 0.150) x (1337 - 333)
[tex]m_{I}[/tex] x 0.461 x 400 = (2.24 + 2.25) x 1004
[tex]m_{I}[/tex] x 184.4 = 4507.96
[tex]m_{I}[/tex] = [tex]\frac{4507.96}{184.4}[/tex]
[tex]m_{I}[/tex] = 24.4466
[tex]m_{I}[/tex] ≅ 24.45
The mass of Iron is 24.45 kg.
A train that is 268.7 m long undergoes constant acceleration the moment the last car (end of
the train) is outside of the station, how far is the front of the train from the station after 25.0 s
if its initial speed before acceleration is 4.48 m/s and its final speed is 27.4 meters per second?
Answer:
its 667m just took the test
Explanation:
How does the use of vectors
allow you to compare,
translate, and determine the
motion of a object?
Answer:
Explanation:
A vector is any substance that has both magnitude and direction. lndeed a vector quantity has both magnitude and direction. Example of a vector quantity is velocity. Determining the velocity of an object allows one to determine how fast and in what direction the object moves. Velocity also affords the opportunity to compare the motion of more than one vectors.
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT, then the rod’s length changes by ΔL=????L0ΔT, where ???? is the thermal expansion coefficient. For steel, ????=1.24×10−5∘C−1. (a) A steel rod has length L0=70cm at T0=70∘C. What is its length at T=110∘C?
Answer:
[tex]\Rightarrow L=70 .03472 cm[/tex]
Explanation:
For convenience, let's represent the thermal expansion coefficient by [tex]\alpha[/tex], i.e. [tex]????=\alpha[/tex].
Given that, for steel [tex]\alpha =1.24\times 10^{-5}[/tex] °[tex]C^{-1}[/tex],
initial length, [tex]L_0=70 cm[/tex], initial temperature, [tex]T_0=70[/tex] °[tex]C[/tex], and the final temperature, [tex]T=110[/tex] °[tex]C[/tex].
Let the length of the rod at [tex]T=110[/tex] °[tex]C[/tex] be [tex]L[/tex].
Now, change in length, [tex]\Delta L=\alpha L_0 \Delta T[/tex]
[tex]\Rightarrow \Delta L=\alpha L_0 (T-T_0)[/tex]
[tex]\Rightarrow L-L_0=1.24\times 10^{-5}\times 70 (110-70)[/tex]
[tex]\Rightarrow L-70=1.24\times 10^{-5}\times 70 \times 40[/tex]
[tex]\Rightarrow L=70 + 0.03472 cm[/tex]
[tex]\Rightarrow L=70 .03472 cm[/tex]
Hence, the length of the rod at [tex]T=110[/tex] °[tex]C[/tex] be [tex]70.03472 cm[/tex].
Find the average velocity (in m/s) of a cyclist that starts 150 meters north of town and is 1200 meters north of town after 30 minutes
Answer:
v = 0.58 m/s
Explanation:
The velocity of an object is given as the ratio of the total distance traveled by the object to the time interval taken for traveling. Hence, we use the following formula to find the average velocity of the cyclist:
v = s/t
where,
v = average velocity = ?
s = distance traveled = final position - initial position =1200 m - 150 m =1050 m
t = time interval = (30 min)(60 s/1 min) = 1800 s
Therefore,
v = 1050 m/1800 s
v = 0.58 m/s
A young child cannot understand
another person's perspective because he is
Answer:
egocentric
Explanation:
A reversible heat engine, operating in a cycle, withdraws thermal energy from a high-temperature reservoir (the temperature of which consequently decreases), performs work w, and rejects thermal energy into a low-temperature reservoir (the temperature of which consequently increases). The two reservoirs are, initially, at the temperatures T1 and T2 and have constant heat capacities C1 and C2, respectively. Calculate the final temperature of the system and the maximum amount of work which can be obtained from the engine.
Answer:
The final temperature is [tex]\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex]
and the maximum amount of workdone is [tex]C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex].
Explanation:
Assume that [tex]R_1[/tex] is the reservior having temperature [tex]T_1 K[/tex] and heat capicity [tex]C_1 \frac JK[/tex] and [tex]R_2[/tex] is the reservior having temperature [tex]T_2[/tex] and heat capicity [tex]C_2 \frac JK[/tex].
The work will be extracted till that both the reservior reach the thermal equilibrium. Let the final temperature of both the reservior is [tex]T_f[/tex].
Let total [tex]Q_1[/tex] heat is extracted by the heat engine from the reservior [tex]R_1[/tex] and its temperature decreases from [tex]T_1[/tex] to [tex]T_f[/tex] and [tex]Q_2[/tex] heat is rejected by the heat engine to the reservior [tex]R_2[/tex] and its temperature decreases from [tex]T_2[/tex] to [tex]T_f[/tex].
So, The maximum amount of work done,
[tex]w= Q_1 - Q_2\; \cdots (i)[/tex]
Now, as the heat engine is reversible, so change is entropy for the universe is 0, which means sum of change in entropy for the ststem as well as surrounding is 0.
As shown in figure, the system is the reversible engine, so, change is entropy for the system is 0. Hence, change in entropy for the the surrounding is 0.
As temperature of [tex]R_1[/tex] is changing fron [tex]T_1[/tex] to [tex]T_f[/tex], so, change in entropy of surrounding due to transfer of [tex]Q_1[/tex] is [tex]C_1 \ln \frac {T_f}{T_1}[/tex].
Similarly, change in entropy of surrounding due to transfer of [tex]Q_2[/tex] is [tex]C_2 \ln \frac {T_f}{T_2}[/tex].
As the net change in entropy of the surrounding is 0.
[tex]\Rightarrow C_1 \ln \frac {T_f}{T_1}+C_2 \ln \frac {T_f}{T_2}=0[/tex]
[tex]\Rightarrow \ln \left( \frac {T_f}{T_1} \right)^{C_1}+ \ln \left( \frac {T_f}{T_2}\right)^{C_2}=0[/tex]
[tex]\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}=- \ln \left( \frac {T_f}{T_2}\right)^{C_2}[/tex]
[tex]\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}= \ln \left( \frac {T_2}{T_f}\right)^{C_2}[/tex]
[tex]\Rightarrow \left( \frac {T_f}{T_1}\right)^{C_1}=\left( \frac {T_2}{T_f}\right)^{C_2}[/tex][taking anti-log both the sides]
[tex]\Rightarrow T_f^{(C_1 +C_2)}=T_1^{C_1}+T_2^{C_2}[/tex]
[tex]\Rightarrow T_f=\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}\; \cdots (ii)[/tex]
This is the required final temperature.
Now, from equarion (i), the maximum amount of work done is
[tex]w= Q_1 - Q_2[/tex]
As [tex]Q=C\Delta T[/tex]
[tex]\Rightarrow w=C_1(T_1-T_f)-C_2(T_f-T_2)[/tex]
[tex]\Rightarrow w=C_1T_1 + C_2T_2-(C_1+C_2)T_f[/tex]
From equation [tex](ii)[/tex],
[tex]w=C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex]
This is the required maximum workdone.
0.00032 cm is equal to
the answer is c because you have to move the decimal back because of the negative
Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at 10.0∘C. What volume of radiator fluid will overflow when the radiator and fluid reach their 95.0∘C operating temperature, given that the fluid’s volume coefficient of expansion is 600×10−6/∘C? Note that this coefficient is approximate, because most car radiators have operating temperatures of greater than 95.0∘C.
Answer:
ΔV = 0.816 L
Explanation:
The change in volume of the fluid upon heating is given by the following formula:
ΔV = βVΔT
where,
ΔV = Increase in Volume of Fluid = Volume of Overflow = ?
β = Coefficient of volumetric expansion of fluid = 600 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature = Final Temperature - Initial Temperature
ΔT = 95°C - 10°C = 85°C
Therefore,
ΔV = (600 x 10⁻⁶ °C⁻¹)(16 L)(85° C)
ΔV = 0.816 L
It's important to match your exercise shoes with the type of exercise in which you will be participating, Please select the best answer from the choices provided OT
Answer:
the answer is true
Explanation:
Fill in the blank
A pot of water is placed on the cook top. For a while the temperature of the water increases,
indicating the____
is increasing.
Answer:
temperature
Explanation:
when you put a water on the stove the water will start to boil there for temperature
1) A plane's velocity increases from 40 m/s to 100 m/s over a 10 second interval. What is the plane's average acceleration for this interval? *
Answer:
average acceleration = 6 [tex]\frac{m}{s^2}[/tex]
Explanation:
Recall that the average acceleration [tex](a)[/tex] is defined by the change in velocity from an initial velocity [tex](v_i)[/tex], to a final velocity [tex](v_f)[/tex] over the time (t) it took that change to happen. Then, in mathematical terms this is:
[tex]a=\frac{v_f-v_i}{t}[/tex]
with our information this becomes:
[tex]a=\frac{v_f-v_i}{t} = \frac{100-40}{10}=6\,\frac{m}{s^2}[/tex]
Two red blood cells each have a mass of 4.60×10−14 kg4.60×10−14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. Once cell carries −2.00 pC−2.00 pC of charge and the other −2.90 pC−2.90 pC , and each cell can be modeled as a sphere 8.20 μm8.20 μm in diameter. What minimum relative speed vv would the red blood cells need when very far away from each other to get close enough to just touch? Ignore viscous drag from the surrounding liquid.
Answer:
v = 5.26 10² m / s
Explanation:
We can solve this exercise using the concepts of conservation of mechanical energy, because there is no friction
starting point. Red blood cells too far away
Em₀ = K = ½ m v²
final point. Red blood cells touching r = 8.20 10⁻⁶ m
Em_f = U = k q₁ q₂ / r₁₂
Em₀ = Em_f
½ m v² = k q₁ q₂ / r₁₂
v = √ (2 k q₁ q₂ / m r₁₂)
we calculate
v =√ (2 9 10⁹ 2 10⁻¹² 2.9 10⁻¹² / (4.60 10⁻¹⁴ 8.20 10⁻⁶))
v = √ (0.276775 10⁶)
v = 0.526 10³ m / s
v = 5.26 10² m / s
If you hit a nail that has a mass of .003 kg with a force of 300 N, what will
the acceleration of the nail be?
Answer:
Explanation:
M= 0.003 kg
F=300N
a=?
F=mass*acceleration
a=F/m
a=300/0.003
a=100,000m/s^2
What are 1A, 3B, and 7A examples of on the periodic table?
groups.
numbers
periods
rows
Answer:
groups
Explanation:
I got a 100 on my quiz
Answer:
It's groups
Explanation:
2020 quiz
20-ohm resistor is connected to a 10 V battery. The battery is then replaced by a battery that provides a larger voltage what happens to the current through the resistor
Explanation:
It is given that,
Resistance, R = 20 ohms
Voltage of the battery, V = 10 V
We can find current flowing through the circuit using Ohm's law as follows :
V = IR
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{10}{20}\\\\I=0.5\ A[/tex]
It can be seen from the Ohm's law, that the current is directly proportional to the voltage. It means that if the battery is replaced by a battery that provides a larger voltage, the current through the circuit will be more than 0.5 A i.e. it increases.
The current via the circuit will be more than 0.5 A i.e. it increases.
Ohm law:Since
Resistance, R = 20 ohms
The voltage of the battery, V = 10 V
Now we applied the above law i.e.
V = IR
I = V/R
I = 10/20
I = 0.5 A
Based on the above calculations, we can say that the current should be directly proportional to the voltage. That means in the case when the battery should be replaced by the battery that gives the larger voltage so the current should be more than 0.5 A due to this it increased.
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10. An object accelerates 2.0 m/s2 when a force of 12.0 newtons is applied
to it. What is the mass of the object?|
Explanation:
F = ma
m = F/a
= 12N / 2m/s2
= 6kg
The mass of the object is 6kg.
What is Force?A force is defined as an effect that can change the motion of an object by which an object with mass can change its velocity, that is, accelerate. Force can also be described simply as a push or pull. A force has both magnitude and direction which makes it a vector quantity.
According to the Newton's second law of motion is F = ma, or force is equal to mass times acceleration.
Where, F= Force applied
m= mass of the object
a= acceleration
For above given information,
F= 12 N
Acceleration= [tex]2 m/s^2[/tex]
So, mass will be m=F/a
m= 12/2 = 6kg
Thus, the mass of the object is 6kg.
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How does the mass of an object affect its motion through the air?
The motion of an object through the air does not affect by its mass. The rate of fall of objects does not depend upon the mass.
What are free fall and air resistance?Free fall is a motion of a body in which gravity is the only force acting upon it. An object moving upwards might not be considered to be falling. But if the object is under the effect of the force of gravity, it is said to be in free fall.
Free fall is a type of motion in which the force acting upon an object is only gravity. Objects are not encountering a significant force of air resistance as they are only falling under the sole influence of gravity. All objects under such conditions will fall with the same rate of acceleration, regardless of their masses.
As an object falls through the air, have gone through some degree of air resistance. Air resistance is the collisions of the object's leading surface with molecules present in the air. The two most common factors that have a direct effect on the amount of air resistance are the cross-sectional area of the object and the speed of the object.
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Pea plants have 2 advantages as genetic specimens:
Answer:
it can be cross pollinated as well as self pollinated
it has short life style
pea plant has many contrasting character in pair example tall, short.
large number of offspring are produced from hybrid plants
A boy walked 22 m East then 43 m West , and then 3 meters East. What is the sum of the boys motion.
Given :
A boy walked 22 m East then 43 m West , and then 3 meters East.
To Find :
What is the sum of the boys motion.
Solution :
We need to find the sum of all the motion of boy .
So , we need to find the distance covered by the boy .
We know , total distance covered is absolute sum of all the motion .
[tex]D=22+43+3\\\\D=68\ meters[/tex]
Therefore , total distance covered is 68 meters .
Hence , this is the required solution .
Answer:
It's 18m West.
Explanation:
Imagine you have a number line. your starting point was 22 east of origin. Then you went 43 meters west which leaves you at 21m west. Then you go back east 3 meters. Which leaves you at 18 m west. I hope this helps.
You carry a fire hose up a ladder to a height of 10 m above ground level and aim the nozzle at a burning roof that is 9 m high. You hold the hose horizontally and notice that the water strikes the roof at a horizontal distance of 7 m from where it exits the nozzle. The hose is connected to a large pressurized chamber in the fire truck 0.5 m above ground level. What is the pressure in the chamber
Answer:
The value is [tex]P_1 = 314645 \ Pa [/tex]
Explanation:
From the question we are told that
The height is [tex]h_2 = 10 m[/tex]
The height of the burning roof is [tex]k = 9 m[/tex]
The horizontal distance is [tex]d = 7 \ m[/tex]
The height of the truck is [tex]h_1 = 0.5 \ m[/tex]
Generally the time for the water to hit the roof from the hose is mathematically represented as
[tex]t = \sqrt{\frac{2 * (h_2 - k)}{g} }[/tex]
=> [tex]t = \sqrt{\frac{2 * (10 - 9)}{9.8} }[/tex]
=> [tex]t = 0.4518 \ s [/tex]
Generally the velocity of the water is mathematically evaluated as
[tex]v_2 = \frac{d}{t}[/tex]
[tex]v_2 = \frac{ 7}{0.4518}[/tex]
[tex]v_2 = 15.5 \ m/s [/tex]
Generally from Bernoulli's Equation we have that
[tex]P_1 + \frac{1}{2} v_1^2 * \rho + \rho *g *h_1 = P_2 + \frac{1}{2} v_2^2 * \rho + \rho *g *h_2[/tex]
Here [tex]P_1 [\tex] is pressure in the chamber which we are to calculate , [tex]P_2 [\tex] is the atmospheric pressure with value [tex]P_2 = 101325 \ Pa [\tex] , [tex]v_1 [\tex] is the velocity of the water before it starts flowing with value [tex]v_1 = 0 m/s [\tex] , [tex]\rho [\tex] is the density of water with value [tex]\rho = 1000 \ kg/m^3 [\tex]
So
[tex]P_1 + \frac{1}{2} 0^2 * 1000 + 1000 *9.81 *0.5 = 101325 + \frac{1}{2}* 15.5^2* 1000 + 1000 *9.81 *10[/tex]
[tex]P_1 = 314645 \ Pa [/tex]
Mary throws a baseball straight upward. We can ignore air resistance.
Answer:
Acceleration Remains Constant and Velocity decreases.
Explanation:
A motorcycle patrolman is monitoring traffic from behind a billboard along a stretch of road where the speed limit is 96.0 km/hr. He clocks a motorist at 107 km/hr and decides to give chase and award the driver a speeding ticket. By the time he gets onto the highway and up to his chase speed of 131 km/hr, he is 350 m behind the speeder. Determine the amount of time it takes the patrolman to catch the speeder.
Answer:
The time taken is [tex]t = 52.5 \ s [/tex]
Explanation:
From the question we are told that
The speed limit is [tex]v__{{l}}} = 96.0 \ km/hr = \frac{96 * 1000}{3600} = 26.7 \ m/s[/tex]
The velocity of the motorist is [tex]v_m = 107 \ km/hr = \frac{107 * 1000}{3600} = 29.72 \ m/s[/tex]
The chase speed of the motorcycle patrolman is [tex]v = 131 \ km/hr = \frac{131 *1000}{3600} = 36.39 \ m/s[/tex]
The relative distance between the motorcycle patrolman and the speeder is d= 350 m
Generally the relative speed between the the motorcycle patrolman and the speeder is mathematically represented as
[tex]v_r = v - v_m[/tex]
=> [tex]v_r = 36.39 - 29.72[/tex]
=> [tex]v_r = 6.67 \ m/s [/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{v_r}{d}[/tex]
=> [tex]t = \frac{350}{ 6.67}[/tex]
=> [tex]t = 52.5 \ s [/tex]
Notice that in each conversion factor the numerator equals the denominator when units are taken into account. A common error in dealing with squares is to square the units inside the parentheses while forgetting to square the numbers! QUESTION What is the numerical part of the time conversion factor, apart from units, that would be used to further convert the answer to km/h2
Answer:
he factor for the temporal part 1.296 107 s² = h²
m / s² = 12960 km / h²
Explanation:
This is a unit conversion exercise.
In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.
In this case, it is requested to convert a measure km/h²
Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is m / s² , which is a measure of acceleration. Let's cut this down
the factor for the distance is 1000 m = 1 km
the factor for time is 3600 s = 1 h
let's make the conversion
m / s² (1km / 1000 m) (3600 s / 1h)²
note that as time is squared the conversion factor is also squared
m / s² = 12960 km / h²
the factor for the temporal part 1.29 107 s² = h²
otential difference ΔVΔV exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 2.70×10−20 J2.70×10−20 J of work is required to eject a positive sodium ion (Na+)(Na+) from the interior of the cell, what is the magnitude of the potential difference (in millivolts) between the inner and outer surfaces of the cell?
Answer:
The value is [tex]V =168.75\ millivolt [/tex]
Explanation:
From the question we are told that
The workdone is [tex]W= 2.70 * 10^{-20 } \ J[/tex]
Generally charge on the positive sodium ion is equivalent to the charge on a proton, the value is [tex]e = 1.60 *10^{-19} \ C[/tex]
Generally the potential difference between the inner and outer surfaces of the cell is mathematically represented as
[tex]V = \frac{W}{e}[/tex]
=> [tex]V = \frac{2.70 * 10^{-20 } }{1.60 *10^{-19} }[/tex]
=> [tex]V = 0.16875 \ V[/tex]
converting to millivolt
[tex]V = 0.16875 * 1000 [/tex]
=> [tex]V =168.75\ millivolt [/tex]
Two supports, made of the same material and initially of equal length, are 2.0 m apart. A stiff board with a length of 4.0 m and a mass of 10 kg is placed on the supports, with one support at the left end and the other at the midpoint. A block is placed on the board a distance of 0.50 m from left end. As a result the board is horizontal (that is, the downward force on each support is the same). The mass of the block is:
Answer:
20 kg
Explanation:
Assuming that the board remains horizontal with the unknown mass on it, then F = kx
If we add the vertical forces to zero, we have something like this
2F - (M + 10)g = 0
2F = (M + 10)g, next, divide both sides by 2
F = (M + 10)g/2
Since we were able to sum the moments at the right end of the board to zero, we then proceed to find the unknown mass M
To start, we say, Let the clockwise moment is positive, and so
F * 4 + F * 2 - Mg * 3.5 - 10 * g * 2 = 0
4F + 2F - 3.5Mg - 10 * 2 * g = 0
6F - 3.5Mg - 10 * 2 * g = 0,
Remember from above, we say that
F = M + 10)g/2, now, all we do is substitute it inside this equation
6 * (M + 10)g/2 - 3.5Mg - 10 * 2 * g = 0
3 * (M + 10)g - 3.5Mg - 10 * g * 2 = 0, divide all sides by g(so as to eliminate it)
3 (M + 10) - 3.5M - 10 * 2= 0
3M + 30 - 3.5M - 20 = 0
-0.5M + 10 = 0
0.5M = 10
M = 10/0.5
M = 20 kg
The mass of the block that was placed on the board is; M = 20 kg
Since the two supports are the same, then it means their forces will be the same.
Thus;
Force at left support = F
Force at right support = F
Now, we are told that a block of unknown mass is placed on the stiff board. This means that the sum of the weight of the board and and the block on it will be; (M + 10)g
where M is the mass of the block.
Now, from equilibrium we know that sum of upward forces is equal to sum of downward forces. Thus;
F + F = (M + 10)g
Thus;
2F = (M + 10)g
F = ¹/₂(M + 10)g
Now, taking moments about the right end gives;
(F × 4) + (F × 2) - (M × 3.5)g - (10 × 2)g = 0
6F - 3.5Mg - 20g = 0
Put ¹/₂(M + 10)g for F to get;
6(¹/₂(M + 10)g) - 3.5Mg - 20g = 0
Divide through by g to get;
3(M + 10) - 3.5M = 20
3M + 30 - 3.5M = 20
3.5M - 3M = 30 - 20
0.5M = 10
M = 10/0.5
M = 20 kg
Read more at; https://brainly.com/question/17059265
A runner has an original velocity of 6 m/s and slows to a final velocity of 0 m/s. If the runner covers a
displacement of 12 m while slowing down, how long time) did it take the runner to stop?
Answer:
4 s
Explanation:
Given:
Δx = 12 m
v₀ = 6 m/s
v = 0 m/s
Find: t
Δx = ½ (v + v₀) t
12 m = ½ (0 m/s + 6 m/s) t
t = 4 s
