Answer:
Wavelength
Explanation:
Select all that apply. The spectrum of Star Y is compared to a reference hydrogen spectrum. What can be concluded about Star Y? Reference (normal) Hydrogen Spectrum Star Y I 500 550 600 650 700 750 400 450 9200 OdyssepWare, Ing. Star Y is showing radial motion Star Y is moving away from the Earth Star Y is moving toward the Earth Star Y is showing transverse motion
Answer:
Star Y is showing radial motion
Star Y is moving toward the Earth
Explanation:
Just answered this question on a quiz and got it right.
Star Y is showing radial motion
Star Y is moving towards earth.
What is a spectrum ?Spectrum is defined as the arrangement of radiations emitted by atoms or molecules based on their characteristic wavelength and frequency.
Here,
The spectrum of the star Y is given and is compared to a reference hydrogen spectrum. The range of wavelengths of the star is given. The spectrum of the star shows that the wavelengths are shifted such that from longer wavelengths to shorter wavelength. This phenomenon of shift in wavelengths from higher to lower is known as blue shift. The star shows blue shift that means shifted to shorter wavelength or it can be said as shifted from lower frequency to higher frequency. As a result, it can be concluded that the star Y is moving towards the earth which implies Star Y is showing radial motion.
Hence,
The star Y is showing radial motion.
Star Y is moving towards the earth.
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A Spring is pulled to 10cm and held in place with a force of 500N .what is the spring constant
We are given:
extension of the spring (x) = 10 cm OR 0.1 m
Force applied to keep it in position (F) = 500 N
Solving for the spring constant:
We know that F = kx (when the spring is extended)
replacing the variables with the given values
500 = k * 0.1
k = 5000 N/m
This means that to extend the spring by 1 m, we have to apply a force of 5000 N
Assume you are in the car and the car is moving at a certain speed to
school. Are you at rest or in motion with respect to the school? With
respect to the car?
You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged
that your client, Mr. Lawton, shot the victim, Mr. Cray. The detective who investigated the scene of the
crime, Mr. Dibny, found a second bullet, from a shot that missed Mr. Cray, that had embedded itself into a
chair. You arise to cross-examine the detective.
You: In what type of chair did you find the bullet?
Dinby: A wooden chair.
You: How massive was this chair?
Dinby: It had a mass of 20 kg.
You: How did the chair respond to being struck with a bullet?
Dinby: It slid across the floor.
You: How far?
Dinby: Three centimeters. The slide marks on the dusty floor are quite distinct.
You: What kind of floor was it?
Dinby: A wood floor.
You: What was the mass of the bullet you retrieved from the chair?
Dinby: Its mass was 10 g.
You: Have you tested the gun you found in Mr. Lawton's possession?
Dinby: I have.
You: What is the muzzle velocity of bullets fired from that gun?
Dinby: The muzzle velocity is 450 m/s.
With only slight hesitation, you turn confidently to the jury and proclaim, "My client's gun did not fire those
shots!"
(a) How are you going to convince the jury and judge?
(b) Choose one part of your solution and perform a sense-making analysis. Clearly state which sensemaking analysis you’ve chosen and why.
Answer:
It was not fired from the client's gun because the chair slid only 3 centimeters . If it had been fired from the client's gun the chair would slid 25.82 centimeters.
Explanation:
According to the law of conservation of momentum the momentum of the system before collision must be equal to the momentum of the system after the collision.
M1u1= m2u2
Let M1 = mass of the chair = 20kg
m2= mass of the bullet= 10g= 0.001kg
u1= velocity of the chair before collision = zero m/s
u2 = velocity of the bullet before collision = zero m/s
v1= velocity of the chair after collision = ? m/s
v2 = velocity of the bullet after collision = 450 m/s
After collision their velocities change from u1 to v1 and u2 to v2 so
M1v1= m2v2
v1= m2v2/M1
v1= 0.01 *450/ 20= 0.225 m/s
Now according to the law of conservation of energy the energy of the system before collision must be equal to the energy of the system after the collision.
The energy of the chair after the bullet is hit is
KE of the chair + KE of the bullet= 1/2 (M)(v1)²+ 1/2 m(v2)²=
1/2 ( 20) (0.225 )² + 1/2 (0.01) (450)²
= 0.50625 + 1012.5= 1013.00625 Joules
Frictional force = Coefficient of kinetic force of wood on wood ( M+m) g
= 0.2* ( 20.01) 9.8= 39.2196 N
Work done by friction = frictional force * distance
If law of conservation of energy is applied the KE must be equal to the work done
KE = W
W= f*d
KE= F*d
d = KE/f= 1013.00625/ 39.2196= 25.82 cm
The chair did not move 25.82 cm .
It only moved 3 centimeter.
Hence the bullet fired was not from the client's gun.
A 5 kg ball and a 2 kg ball are placed at opposite ends of a massless beam so that the system is in equilibrium. What is the ratio of the length of the beam to the distance from the heavier ball to the pivot
Answer:
L/x = 3.5 (The ratio of length of beam to the distance from heavier ball to pivot).
Explanation:
In order for the system to be in equilibrium, the moment created by both masses about the pivot point must be equal:
m₁x = m₂y
where,
m₁ = 5 kg
m₂ = 2 kg
x = distance of 5 kg ball from pivot
y = distance of 2 kg ball from pivot
Therefore,
(5 kg)x = (2 kg)y
y = (5kg/2kg)x
y = 2.5 x
but,
x + y = L
where,
L = length of beam
using the value of y from the previous equation:
x + 2.5 x = L
3.5 x = L
L/x = 3.5 (The ratio of length of beam to the distance from heavier ball to pivot).
Forces that are equal in size but opposite in direction are called what ?
Answer:
I think they are called balanced forces
Explanation:
A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
Answer:
Explanation:
To get the focal length, we will use the lens formula;
1/f = 1/u + 1/v
f is the focal length
u is the object distance
v is the image distance
Given
since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.
u = 35cm
v = -2.0
1/f = 1/35-1/2
1/f = 2-35/70
1/f = -33/70
f = -70/33
f = -2.12 cm
f = -0.0212m
Power of a lend is the reciprocal of its focal length
Power of the lens = 1/f
P = 1/-0.0212
P = -47.17dioptres
The power of the lens is -47.17D
Find the magnitude of the gravitational force (in N) between a planet with mass 6.00 X 10^24 kg and its moon, with mass 2.50 X 10^22 kg, if the average distance between their centers is 2.70 X 10^8 m. What is the moon's acceleration (in m/s2) toward the planet? (Enter the magnitude.) What is the planet's acceleration (in m/s2) toward the moon? (Enter the magnitude.)
Answer:
Magnitude of gravitational force between this planet and its moon: approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Acceleration of this moon towards the planet: approximately [tex]5.49 \times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
Acceleration of this planet towards its moon: approximately [tex]2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
Explanation:
Look up the gravitational constant, [tex]G[/tex]:
[tex]G \approx 6.67 \times 10^{-11}\; \rm m^3\cdot kg^{-1} \cdot s^{-2}[/tex].
Assume that both this planet and its moon are spheres of uniform density. When studying the gravitational interaction between this planet and its moon, this assumption allows them to be considered as two point masses.
The formula for the size of gravitational force between two point masses [tex]m_1[/tex] and [tex]m_2[/tex] with a distance of [tex]r[/tex] in between is:
[tex]\displaystyle F = \frac{G \cdot m_1 \cdot m_2}{r^2}[/tex],
where [tex]G[/tex] is the gravitational constant.
Let [tex]m_1[/tex] and [tex]m_2[/tex] denote the mass of this planet and its moon, respectively.
Calculate the size of gravitational force between this planet and its moon:
[tex]\begin{aligned} F &= \frac{G \cdot m_1 \cdot m_2}{r^2} \\ &\approx \frac{6.67 \times 10^{-11}\; \rm m^3 \cdot kg^{-1} \cdot s^{-2} \times 6.00 \times 10^{24}\; \rm kg \times 2.50 \times 10^{22}\; \rm kg}{{\left(2.70 \times 10^{8}\; \rm m\right)}^2} \\ &\approx 1.37 \times 10^{20}\; \rm N\end{aligned}[/tex].
Assume that other than the gravitational force between this planet and its moon, all other forces (e.g., gravitational force between this planet and the star) are negligible. The magnitude of the net force on the planet and on the moon should both be approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Apply Newton's Second Law of motion to find the acceleration of this planet and its moon:
[tex]\displaystyle \text{acceleration} = \frac{\text{net force}}{\text{mass}}[/tex].
For this moon:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{2.50 \times 10^{22}\; \rm kg} \approx 5.49\times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
For this planet:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{6.00 \times 10^{24}\; \rm kg} \approx 2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
If you are riding your bike west to your friend’s house, and you ride the 1.25 miles in 5 minutes, what is your velocity, in miles per hour?
Answer:
velocity = 15 miles / hourExplanation:
distance = 1.25 mile
time traveled = 5 min.
find velocity in miles / hour
solution:
use the formula: velocity = distance / time
velocity = 1.25 mile x 60 min
5 min 1 hour
velocity = 15 miles / hour
A typical blood pressure is 120 mm Hg. How high would you need to hang an IV bag so that the fluid enters the blood stream at this pressure? Assume the IV fluid has a density of 1000 kg/m^3.
Answer:
1.63 m
Explanation:
The following data were obtained from the question given above:
Pressure (P) = 120 mmHg
Density (d) = 1000 Kg/m³
Height (h) =?
Next, we shall convert 120 mmHg to N/m². This can be obtained as follow:
760 mmHg = 101325 N/m²
Therefore,
120 mmHg = 120 mmHg × 101325 N/m² / 760 mmHg
120 mmHg = 15998.68 N/m²
Thus, 120 mmHg is equivalent to 15998.68 N/m².
Finally, we shall determine the height to which the IV bag should be placed so that the fluid can enter the blood stream. This is illustrated below:
Pressure (P) = 15998.68 N/m².
Density (d) = 1000 Kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
P = dgh
15998.68 = 1000 × 9.8 × h
15998.68 = 9800 × h
Divide both side by 9800
h = 15998.68 / 9800
h = 1.63 m
Therefore, the the IV bag should be placed at a height of 1.63 m
How many stars are in the universe (approximately)? O 40 sixtillion 0 365 billion O 86.4 million O one
Answer:
a i belive
Explanation:
the univerce is VERY large so a, if im wrong i apologise :(
Find the density of a liquid, in lb/ft^3, that exerts a pressure of 0.400 lb/in^2 at a depth of 42.0 in.
Given :
Pressure, P = 0.4 lb/in².
Depth, h = 42 in.
To Find :
The density of a liquid.
Solution :
Pressure at height h of a liquid of density [tex]\rho[/tex] is given by :
[tex]P = \rho g h[/tex] ....1)
Here, g = 386.04 in/s²( acceleration due to gravity )
Putting all values in equation 1, we get :
[tex]P = \rho g h\\\\0.4 = \rho\times 386.04 \times 42\\\\\rho=\dfrac{0.4}{42\times 386.04}\ lb/in^3\\\\\rho=\dfrac{0.4}{42\times 386.04}\times 12^3\ lb/ft^3\\\\\rho=0.0426\ lb/ft^3[/tex]
Hence, this is the required solution.
which one is correct:
Average acceleration of an object is the:
Change in its position divided by the change in time
Change in time divided by the change in its position
Change in its velocity divided by the change in time
Change in time divided by the change in its velocity
Answer:
Option (c) is correct.
Explanation:
Acceleration of an object is given by the formula as follows :
[tex]a=\dfrac{v-u}{t}[/tex]
Where
u and v are initial and final velocity
t is time
(v-u) is also called the change in velocity
So, the acceleration of an object is equal to the rate of change of velocity. Hence, the correct option is (c) " Change in its velocity divided by the change in time".
The CERN particle accelerator is circular with a circumference of 7.0 km.
Required:
a. What is the acceleration of the protons (m=1.67×10^−27kg) that move around the accelerator at of the speed of light? (The speed of light is v=3.00×10^8m/s.)
b. What is the force on the protons?
Answer:
Explanation:
a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;
a = mv²/r
m is the mass of proton = 1.67×10^−27kg
v is the velocity = 3.00×10^8m/s
r is the radius
Since C = 2πr
7000m = 2πr
r = 7000/2π
r = 1114.08m
Substitute
a = 1.67×10^−27 (3.00×10^8)²/1114.08
a = 1.67×10^−27 * 9×10^16/1114.08
a = 15.03*10^-11/1114.08
a = 0.001346*10^-11
a = 1.346*10^-14m/s²
b) Force on the proton = mass * acceleration
Force = 1.67×10^−27kg * 1.346*10^-14
Force = 2.246*10^-41N
hence the force on the proton is 2.246*10^-41N
2. How much time will it take to move 120 m if an object
moves 300 m/s?
Answer:
2.5 seconds
Explanation:
Time is equal to distance over speed
300 divide by 120 u get 2.5 seconds
A 5-newton force directed east and a 5-newton force directed
north act concurrently on a point. Calculate the resultant of the two forces.
Answer:
[tex]5\sqrt{2} \ N[/tex]
Explanation:
It is given that,
A 5-newton force directed east and a 5-newton force directed north.
We need to find the resultant of the two forces.
As the two forces are acting in perpendicular to each other. The resultant of two such forces is given by :
[tex]F=\sqrt{5^2+5^2} \\\\=5\sqrt{2} \ N[/tex]
So, the resulatnt force is [tex]5\sqrt{2} \ N[/tex]
In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of mass 1.67 × 10-27 kg and an orbiting electron of mass 9.11 × 10-31 kg. In one of its orbits, the electron is 5.3 × 10-11 m from the proton. What is the mutual attractive gravitational force between the electron and proton
Answer:
[tex]F=3.61\times 10^{-47}\ N[/tex]
Explanation:
Mass of a proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]
Mass of an electron, [tex]m_e=9.11\times 10^{-31}\ kg[/tex]
The distance between the electron and the proton is, [tex]r=5.3\times 10^{-11}\ m[/tex]
We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :
[tex]F=G\dfrac{m_em_p}{r^2}[/tex]
Where G is the universal Gravitational constant
[tex]F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N[/tex]
So, the force between the electron and proton is [tex]3.61\times 10^{-47}\ N[/tex].
A 30 kg box is being pulled with a force of 125 N. The coefficient of static friction between the box and the floor is 0.35. What is the minimum downward force on the box that will keep it from slipping?
Answer:
The minimum downward force on the box that will keep it from slipping is 63.14 N
Explanation:
Given;
mass of the object, m = 30 kg
applied force, f = 125 N
coefficient of static friction, μ = 0.35
Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.
The net vertical forces on the box is given by;
R - mg - F = 0
F = R - mg
Now, determine normal reaction, R
f = μR
R = f / μ
R = 125 / 0.35
R = 357.14 N
Finally, determine the minimum downward force on the box that will keep it from slipping;
F = R - mg
F = 357.14 - (30 x 9.8)
F = 357.14 - 294
F = 63.14 N
Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N
A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?
Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198
Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]
= [tex]\frac{0.779}{6}[/tex]
= 0.12983
The average coefficient of static friction is 0.130
The average coefficient of static friction is 0.13.
The coefficient of static friction is obtained using the formula; μ = F/R
Where;
F = force acting on the body
R = reaction
μ = coefficient of static friction
The average of measurements is given as; ∑summation of measurements/number of measurements
We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;
0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198/6
= 0.13
The average coefficient of static friction is 0.13
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A good alternative to sit ups is
A. Crunches
B. Leg Raises
C. Side Twists
What unit is used to measure energy when calculating specific heat capacity? Give the abbreviation, not the full name.
The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).
What unit is used to measure energy?The specific heat capacity should be determined in joules per kilogram degree-celsius ( J k g − 1 ∘ C − 1 ).
It is to be supplied for the substance with respect to mass and it increased the temperature.
Hence, The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).
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The unit used to measure energy when calculating specific heat capacity is the joule (J).
The joule (J), a unit used to quantify energy, is used to calculate specific heat capacity. In the International System of Units (SI), the joule serves as the default unit of energy.
It is described as the quantity of energy that is delivered when one newton of force is exerted across a one-meter distance.
Thus, the quantity of heat energy needed to increase the temperature of a particular substance by a specific amount is measured as specific heat capacity. J/kg°C, or joules per kilogramme per degree Celsius, is the unit of measurement.
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A change of temperature of 20 C is equivalent to a change in thermodynamic temperature of
Answer:
20 K
Explanation:
It is given that,
The change in temperature is 20 C.
We need to find the change in thermodynamic temperature.
If teperauture T₁ = 0° C = 0+273 = 273 K
T₂ = 20° C = 20 + 273 = 293 K
The change in temperature,
[tex]\Delta T=T_2-T_1\\\\=293-273\\\\=20\ K[/tex]
So, the change in temperature of 20°C is equivalent to 20 K.
The change in thermodynamic temperature is equal to 20 Kelvin.
Given the following data:
Change in temperature = 20°CTo determine the change in thermodynamic temperature:
A thermodynamic temperature can be defined as an absolute measure of the average total internal energy possessed by a body or an object. Thus, thermodynamic temperature is typically measured in Kelvin (K).
Mathematically, the change in temperature of an object is given by the formula:
[tex]\theta = T_f - T_i[/tex]
Where:
[tex]T_i[/tex] is the initial temperature (0°C)[tex]T_f[/tex] is the final temperature (20°C).Next, we would convert the values in degree Celsius to Kelvin:
Conversion:
0°C = 273 K
20°C = 273 + 20 = 293 K
For the change in thermodynamic temperature:
[tex]\theta = 293 - 273\\\\\theta =20\;K[/tex]
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Draw the resultant of each set of vectors. Indicate if you used the parallelogram or head-to-tail method.
Answer:
Explanation:
See attached file.
an a astronaut stands on the surface of a spherical asteroid that has a weak gravitational field but no atmosphere
Answer:
so what is your question that's not a question
An asteroid is a minor planet in the inner solar system. The asteroid orbits the sun and millions of asteroids exit as remains of planetesimals.
The weak gravitational field of the asteroid describes that there is no contact between the asteroid and the astronaut. As the asteroid has no atmosphere the astronaut jumps and drifts.Hence there is no contact as no gravitational force.
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Describe which relationships in Ohm’s law are DIRECT and which are INVERSE. Use examples to support your answer. You can use calculations, drawings, or graphs to make your point more clear
Answer:
I like to memorize excerpt from articles to solve and answer questions like these. I hope this can help, it's from study.com: "The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r."
C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
Answer:
The mass of the cart is 150 kg.
Explanation:
Given that,
Mass of a boy, m₁ = 50 kg
Initial speed of boy, u₁ = 10 m/s
Initial speed of car, u₂ = 0 (at rest)
The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s
Let m₂ is the mass of the cart. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]
So, the mass of the cart is 150 kg.
1.) There was an earthquake in Salt Lake City, Utah, on March 18, 2020, in the morning at 9 hours, 9 minutes, and 45 seconds Mountain Standard Time (9:9:45 MST). If the velocity of the p-wave is 7.3 km/sec and the velocity of the s-wave is 5.1 km/sec and the s-p time lag is 16 seconds, what is the distance in kilometers from Salt Lake City to the focus of the earthquake? Explain how you calculated the answer.
Answer:
7 because salt lake and Southis weat
A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 1/160 of its mass as exhaust gas and has an acceleration of 14.0 m/s^2.
Required:
What is the speed vgas of the exhaust gas relative to the rocket?
Answer:
840 m/s
Explanation:
Given that,
In the first second the rocket ejects 1/160 of its mass as exhaust gas and has an acceleration of 14.0 m/s².
We need to find the speed of the exhaust gas relative to the rocket.
The thrust of rocket is given by :
[tex]T=v_{gas}\dfrac{dm}{dt}\\\\ma=v_{gas}\dfrac{dm}{dt}\\\\v_{gas}=\dfrac{ma}{\dfrac{dm}{dt}}\\\\v_{gas}=\dfrac{14m}{\dfrac{1}{60}m}\\\\v_{gas}=840\ m/s[/tex]
So, the speed of the exhaust gas relative to the rocket is 840 m/s.
The speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s
Given data :
In first round Rocket ejects 1/60 of mass as exhaust gas
Acceleration of rocket ( a ) = 14.0 m/s²
Determine the speed of the exhaust gas relative to rocket
We will apply the equation for Rocket thrust
T = Vgas * [tex]\frac{dm}{dt}[/tex]
where : T = ma
∴ Vgas = ma / [tex]\frac{dm}{dt}[/tex]
= 14 m / [tex]\frac{1}{60}[/tex] m
therefore V gas = 840 m/s
Hence we can conclude that the speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s
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PLEASE PROVIDE AN EXPLANATION
THANK YOU!
Answer:
(a) 0.993 s
(b) 14.0 N/m
(c) -3.02 m/s
(d) -6.01 m/s²
Explanation:
(a) The block's position can be modeled as a cosine wave:
x(t) = A cos(ωt)
where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.
At t = 0.200 s, x(t) = 15.0 cm.
15.0 cm = 50.0 cm cos((0.200 s) ω)
0.3 = cos((0.2 s) ω)
1.266 rad = (0.2 s) ω
ω = 6.33 rad/s
The period is:
T = (2π rad) (1 s / 6.33 rad)
T = 0.993
(b) For a spring-mass system, ω = √(k/m). The mass of the block is 0.350 kg, so:
ω = √(k/m)
6.33 rad/s = √(k / 0.350 kg)
6.33 rad/s = √(k / 0.350 kg)
40.1 rad/s² = k / 0.350 kg
k = 14.0 N/m
(c) Energy is conserved:
EE₀ = EE + KE
½ kx₀² = ½ kx² + ½ mv²
kx₀² = kx² + mv²
(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²
v = -3.02 m/s
Alternatively, we can take the derivative of our position equation:
v(t) = -Aω sin(ωt)
v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))
v = -3.02 m/s
(d) Sum of forces on the block:
∑F = ma
-kx = ma
a = -kx / m
a = -(14.0 N/m) (0.15 m) / (0.350 kg)
a = -6.01 m/s²
Alternatively, we can take the derivative of our velocity equation:
a(t) = -Aω² cos(ωt)
a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))
a = -6.01 m/s²
we feel cold in winter when we come out from the quilt but the same room becomes warmer after coming back from outside the room
Answer:
Yes
Explanation:
I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.
HOPE THIS HELPED