Answer:
[tex]T_2=630^{\circ}C[/tex]'
Explanation:
Original temperature of the gas, [tex]T_1=28^{\circ}C=301\ K[/tex]
From the ideal gas equation,
[tex]P_1V_1=nRT_1[/tex]
Since,
[tex]P_2=3P_1[/tex]
[tex]nRT_2=3(nRT_1)[/tex]
[tex]T_2=3T_1[/tex]
[tex]T_2=3\times 301[/tex]
[tex]T_2=903\ K[/tex]
or
[tex]T_2=630^{\circ}C[/tex]
So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.
If the radius of a blood vessel drops to 80.0% of its original radius because of the buildup of plaque, and the body responds by increasing the pressure difference across the blood vessel by 10.0%, what will have happened to the flow rate? The flow rate will have changed to _______ % of its original value.
Answer:
40.96%
Solution:
As per the question:
If the original radius be 'r' and the initial pressure difference be 'P':
After the drop, radius:
r' = 0.8r
After rise in pressure:
P' = 0.1 P
Now,
Rate of flow is given by:
R ∝ [tex]r^{4}P[/tex]
Thus
[tex]\frac{R'}{R} = (\frac{0.8r}{r})^{4}\frac{0.1P}{P}[/tex]
R' = 40.96%
A 2-kg wood block is pulled by a string across a rough horizontal floor. The string exerts a tension force of 30 N on the block at an angle of 20º above the horizontal. The block moves at constant speed. If the block is pulled for a distance of 3.0 m, how much work is done by the tension force?
Answer:
Work done, W = 84.57 Joules
Explanation:
It is given that,
Mass of the wooden block, m = 2 kg
Tension force acting on the string, F = 30 N
Angle made by the block with the horizontal, [tex]\theta=20^{\circ}[/tex]
Distance covered by the block, d = 3 m
Let W is the work done by the tension force. It can be calculated as :
[tex]W=F\ cos\theta\times d[/tex]
[tex]W=30\times cos(20)\times 3[/tex]
W = 84.57 Joules
So, the work done by the tension force is 84.57 Joules. Hence, this is the required solution.
The quantity of work done by this tension force is equal to 84.57 Joules.
Given the following data:
Mass of wooden block, m = 2 kg
Tension force, F = 30 N
Angle = 20°
Distance, d = 3 m.
How to calculate the quantity of work done?In Science, work done is generally calculated by multiplying tension force and the vertical distance experienced by an object.
Mathematically, this is given by:
W = Fdcosθ
W = 30 × 3.0 × cos20
Work done, W = 84.57 Joules.
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A 0.72-m long string has a mass of 4.2 g. The string is under a tension of 84.1 N. What is the speed of a wave on this string?
The speed of the wave in the string is 83.4 m/s
Explanation:
For a standing wave in a string, the speed of the wave is given by the equation:
[tex]v=\frac{1}{2L}\sqrt{\frac{T}{m/L}}[/tex]
where
L is the length of the string
T is the tension in the string
m is the mass of the string
In this problem, we have:
L = 0.72 m
m = 4.2 g = 0.0042 kg
T = 84.1 N
Solving the equation, we find the speed of the wave:
[tex]v=\frac{1}{2(0.72)}\sqrt{\frac{84.1}{0.0042/0.72}}=83.4 m/s[/tex]
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A cannon fires a cannonball of mass 16.0 kg by applying a force of 2750 N along the 1.25 m length of the barrel. (a) How much work does the cannon do on the cannonball? (b) The cannon is aimed at a 25.0° angle above the horizontal. Assume gravity is the only other force acting on the cannonball as it moves through the cannon barrel. (That is, ignore all frictional forces.) What is the net work done by these 2 forces on the cannonball while it is in the cannon barrel?
To develop this problem it is necessary to apply the concepts related to Work and energy conservation.
By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,
[tex]W = F*d[/tex]
Where,
F= Force
d = Distance
On the other hand we know that the potential energy of a body is given based on height and weight, that is
[tex]PE = -mgh[/tex]
The total work done would be given by the conservation and sum of these energies, that is to say
[tex]W_{net} = W+PE[/tex]
PART A) Applying the work formula,
[tex]W = F*d\\W = 2750*1.25\\W = 3437.5J[/tex]
PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then
[tex]PE = -mgh*sin25[/tex]
[tex]PE = -16*9.81*1.15sin25[/tex]
[tex]PE = -82.9J[/tex]
The net work would then be given by
[tex]W_{net} = W+PE[/tex]
[tex]W_{net} = 3437.5J-82.9J[/tex]
[tex]W_{net} = 3354.6J[/tex]
Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is 11.8 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.
To solve this problem it is necessary to apply the concepts based on Newton's second law and the Centripetal Force.
That is to say,
[tex]F_c = F_w[/tex]
Where,
[tex]F_c =[/tex]Centripetal Force
[tex]F_w =[/tex]Weight Force
Expanding the terms we have to,
[tex]mg = \frac{mv^2}{r}[/tex]
[tex]gr = v^2[/tex]
[tex]v = \sqrt{gr}[/tex]
Where,
r = Radius
g = Gravity
v = Velocity
Replacing with our values we have
[tex]v = \sqrt{(9.8)(11.8)}[/tex]
[tex]v = 10.75m/s[/tex]
Therefore the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top is 10.75m/s
To get total internal reflection at the interface of water (refractive index 1.33) and a plastic whose refractive index is 1.54:
Which material must the light start in?
A. it doesn't matter
B. water
C. plastic
What is the critical angle?
_____ degrees
Answer:
Explanation:
Total internal reflection refers to a phenomenon that occurs when light travels from a denser medium to less dense medium, in which the incidence ray is inclined at an angle greater than a certain critical angle to the normal ( angle at which the refracted ray is equal to 90°), instead of refraction, the ray is reflected back into the material.
since it must start from a denser medium or a medium with higher refractive index, then it must start with plastics
to calculate critical angle,
we use Snell's law
n₁ sin C = n₂ sin r since r is 90° where n₁ =1.54, n₂ = 1.33
C = sin⁻¹ ( n₂ / n₁) = sin⁻¹ ( 1.33 / 1.54) = 59.7°
the critical angle = 59.7°
The light must start in the plastic for total internal reflection to occur. The critical angle, which is the angle of incidence where the refracted light is tangential to the boundary, is approximately 59.46 degrees.
Explanation:For total internal reflection to occur, the light must go from a material with a higher refractive index to one with a lower refractive index. In this case, the light must start in the plastic.
The critical angle is the angle of incidence for which the angle of refraction equals 90 degrees. It can be calculated using Snell's Law (n1*sin(θ1) = n2*sin(θ2)), where n1 and n2 are the refractive indices and θ1 and θ2 are the angles of incidence and refraction, respectively. If we set θ2 to 90 degrees and rearrange the equation we get: θ1 (critical angle) = arcsin(n2/n1).
Plugging in the given index values, we get: θ1 = arcsin(1.33/1.54), which is approximately 59.46 degrees.
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Light of wavelength 600 nm in vacuum is incident nearly perpendicularly on a thin film whose index of refraction is 1.5. The light travels from the top surface of the film to the bottom surface, reflects from the bottom surface, and returns to the top surface, as the drawing indicates. What is the total (down-and-back) distance traveled by the light inside the film? Express your answer in terms of the wavelength λfilm of the light within the film.
Answer:
6λ_film
Explanation:
Light of wavelength λ 600 nm
Assuming length of thin film of water to be 1200 nm
Total distance traveled = 2×1200 ×10^{-9} = [tex]\frac{2.4\times10^-6}{4\times10^{-7} }[/tex] m
Now, Wavelength of light in film = λ/n
n= refractive index
= [tex]\frac{600\times10^{-9}}{1.5}[/tex] = 4×10^{-7}
No. of wavelengths = distance traveled/Wavelength of light in film
= [tex]\frac{2.4\times10^-6}{4\times10^{-7} }[/tex] = 6
therefore, the total (down-and-back) distance traveled by the light inside the film in terms of wavelength λfilm = 6λfilm
Final answer:
The total distance that the light travels inside the thin film is equal to the wavelength of light within the film (λfilm), as the light travels down and back up, covering a distance of 2t, which is equivalent to λfilm.
Explanation:
When light of wavelength 600 nm in vacuum (represented by λ) is incident on a thin film with an index of refraction of 1.5, we need to calculate the total distance traveled by the light within the film. The wavelength of light within the film (λfilm) is given by the formula λfilm = λ / n, where n is the index of refraction of the film. Given that light hits the film perpendicularly, it travels down to the bottom surface and back up, covering a distance of 2t, where t is the thickness of the film. To express the total distance in terms of λfilm, we use the relationship 2t = 2(λfilm / (2n)) = λfilm since n is already considered in the λfilm. Thus, the total distance traveled by the light inside the film is λfilm.
Suppose that there are N=108 two-state systems, each with energy difference E=6 × 10-21 J between the two states. The environment temperature is 300 K. The number of systems in their high energy states is M. Therefore, the free energy is a function of M: F(M).What is dU/dM?
Answer:
The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].
Explanation:
Given that,
Number [tex]N=10^{8}[/tex]
Energy difference = 6\times10^{-21}\ J[/tex]
Temperature T =300 K
We need to calculate the value of [tex]\dfrac{dU}{dM}[/tex]
We know that,
[tex]\dfrac{dU}{dM}=\dfrac{energy\ difference}{Change\ in\ number\ of\ system}[/tex]
[tex]\dfrac{dU}{dM}=\dfrac{6\times10^{-21}}{10^{8}}[/tex]
[tex]\dfrac{dU}{dM}=6\times10^{-29}\ J/unit[/tex]
Hence, The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].
A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp, measured from the horizontal, at a speed of 40.0 m/s (144 km/h). The top of the ramp is at the same height as the roofs of the buses and each bus is 20.0 m long.
Answer:
Dare devil can cross 7 buses.
Explanation:
given,
angle of inclination of ramp = 32°
speed of the motorcycle = 40 m/s
length of bus = 20 m
how many buses daredevil can clear =?
to solve this we need to calculate the range of the daredevil
considering it as projectile
the range of motorcyclist
[tex]R = \dfrac{V^2 sin (2\theta)}{g}[/tex]
[tex]R = \dfrac{40^2 sin (2\times 32^0)}{9.8}[/tex]
[tex]R =163.26 \times sin(2\times 32^0)[/tex]
[tex]R =146.74\ m[/tex]
length of bus is given as 20 m
Number of bus daredevil can cross
[tex]N = \dfrac{147.74}{20}[/tex]
[tex]N =7.34[/tex]
Dare devil can cross 7 buses.
Answer:
The number of buses are 7.
Explanation:
Given that,
Angle =32°
Speed = 40.0 m/s
Length of bus = 20.0
We need to calculate the range of bus
Using formula of range
[tex]R=\dfrac{v^2\sin2\theta}{g}[/tex]
Where, g = acceleration due to gravity
v = initial velocity
Put the value into the formula
[tex]R=\dfrac{(40.0)^2\times\sin(2\times32)}{9.8}[/tex]
[tex]R=150.20\ m[/tex]
We need to calculate the number of buses
Using formula of number of buses
[tex]N=\dfrac{R}{L}[/tex]
Where, R = range
L = length of bus
[tex]N=\dfrac{150.20}{20.0}[/tex]
[tex]N=7[/tex]
Hence, The number of buses are 7.
Which one of the following statements concerning the electric dipole moment is false?
A) The dipole moment is a scalar quantity.
B) The dipole moment has units of C·m.
C) The dipole moment combines two intrinsic properties of an electric dipole, the distance between the charges and the amount of charge on each object.
D) The dipole moment is directed from the negative charge toward the positive charge of the dipole.
E) The direction of the dipole moment is used to specify the orientation of the dipole.
Answer:
A.The dipole moment is a scalar quantity.
Explanation:
We know that
Dipole moment :It measure the polarity of chemical bond in a molecule .
The electric dipole moment is equal to product of any charge (positive or negative ) and the distance between the two charges.
Mathematical representation:
[tex]\mu=q\times d[/tex]
Where [tex]\mu[/tex] =Dipole moment
q=Charge on atom or particle
d=Distance between two charged particles
It helps to find out the molecule is polar or non- polar.
When the dipole moment is zero then the molecule is non-polar.
Dipole moment is a vector quantity.
The direction of dipole moment from negative charge to positive.
When unit of charge is C and unit of distance is m.
Then, unit of dipole moment=C-m
Hence, option A is false.
Final answer:
The false statement about the electric dipole moment is that it is a scalar quantity; it is a vector quantity. So the correct option is A.
Explanation:
The statement concerning the electric dipole moment that is false is A) The dipole moment is a scalar quantity. This is incorrect because the electric dipole moment is a vector quantity, not a scalar. The correct statements about the dipole moment are that it does indeed have units of C·m (Statement B), it combines both the distance between the charges and the magnitude of the charges (Statement C), it is directed from the negative charge toward the positive charge (Statement D), and its direction specifies the orientation of the dipole (Statement E).
The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air. By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius.
To solve this problem it is necessary to apply the concept related to density and its definition with respect to mass and volume.
Density can be expressed as
[tex]\rho = \frac{m}{V}[/tex]
Where,
m = mass
V = Volume
The average density of fish air system is equal to the neutral bouyancy when the water displaced by the fish air system can be expressed as
[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]
Where,
[tex]m_a[/tex]= Mass of air
[tex]m_f =[/tex] Mass of fish
[tex]V_a[/tex]= Volume of air
[tex]V_f[/tex]= Volume of fish
As we know that m = \rho V we can replacing the equivalent value for the mass fro fish and air, then
[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]
[tex]\rho_{b} = \frac{\rho_a V_a+\rho_f V_f}{V_a+V_f}[/tex]
[tex]\rho_{b}(V_a+V_f)=\rho_a V_a+\rho_f V_f[/tex]
[tex]\rho_{b} V_a+\rho_{b}V_f = \rho_a V_a+\rho_f V_f[/tex]
[tex]\rho_{b} V_a- \rho_a V_a=\rho_f V_f-\rho_{b}V_f[/tex]
[tex]V_a(\rho_{b}-\rho_a)=V_f(\rho_f-\rho_{b})[/tex]
[tex]\frac{V_a}{\rho_f}=\frac{(\rho_f-\rho_{b})}{(\rho_{b}-\rho_a)}[/tex]
Replacing with our values we have that
[tex]\frac{V_a}{\rho_f}=\frac{(1000-1080)}{(1.2-1000)}[/tex]
[tex]\frac{V_a}{\rho_f}=0.08[/tex]
[tex]\frac{V_a}{\rho_f}=8\%[/tex]
Therefore the percentage of Volume is increased by 8%
To achieve neutral buoyancy in fresh water, a fish with an initial density of 1080 kg/m^3 needs to increase its volume by 8%, utilizing its swim bladder.
Explanation:The solution to this problem involves understanding Archimedes' Principle and the concept of density. In order to achieve neutral buoyancy, the fish needs to match its density to the density of freshwater, which is approximately 1000 kg/m³. It can do this by increasing its volume through inflating its swim bladder, thus reducing its overall density.
Since the average density of an object is mass divided by its volume, we can set up an equation involving the initial and final densities of the fish, with the initial density being 1080 kg/m³ and the final density being 1000 kg/m³. Let the initial volume of the fish be V1, the final volume be V2, and the mass of the fish (which doesn't change) be m. Therefore, m/V1 = 1080 and m/V2 = 1000. We can then solve these equations to find that V2 = 1.08V1.
In terms of percent increase, (V2 - V1) / V1 = 0.08 or 8%. Hence, the fish needs to increase its volume by 8% to achieve neutral buoyancy in freshwater.
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a kg object and a kg object are separated by m. find the magnitude of the net gravitational force exerted by these objects on a kg object placed midway between them. at what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?
Answer:
F₂ = 0 N
x = 0.5*d = 0.5 m
Explanation:
Given
m₁ = m₂ = m₃ = m = 1 Kg
d = distance between m₁ and m₃ = d₁₃ = 1 m
d₁₂ = 0.5*d = 0.5*(1 m) = 0.5 m = d₃₂
G = 6.673*10⁻¹¹ N*m²/ Kg²
a) Find the magnitude of the net gravitational force exerted by these objects on a kg object (F₃) placed midway between them.
we can apply
F₂ = F₁₂ + F₃₂
then we use the formula
F₁₂ = G*m₁*m₂ / d₁₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₁₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₁₂ = 2.6692*10⁻¹⁰ N (←)
and
F₃₂ = G*m₃*m₂ / d₃₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₃₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₃₂ = 2.6692*10⁻¹⁰ N (→)
we get
F₂ = F₁₂ + F₃₂ = (- 2.6692*10⁻¹⁰ N) + (2.6692*10⁻¹⁰ N) = 0 N
b) At what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?
From a) It is known that x = 0.5*d = 0.5*(1 m) = 0.5 m
Nevertheless, we can find the distance as follows
If
F₂ = 0 ⇒ F₁₂ + F₃₂ = 0 ⇒ F₁₂ = - F₃₂
If
x = distance between m₁ and m₂
d - x = distance between m₂ and m₃
we get
F₁₂ = G*m² / x²
F₃₂ = G*m² / (d - x)²
If F₁₂ = - F₃₂
⇒ G*m² / x² = - G*m² / (d - x)²
⇒ 1 / x² = - 1 / (d - x)²
⇒ (d - x)² = - x²
⇒ d²- 2*d*x + x² = - x²
⇒ 2*x² - 2*d*x + d² = 0
⇒ 2*x² - 2*1*x + 1² = 0
⇒ 2*x² - 2*x + 1 = 0
Solving the equation we obtain
x = 0.5 m = 0.5*d
An object is at rest on top of a smooth sphere with a radius of ???? = 15.3 m that is buried exactly halfway under the ground. It then begins to slide down. At what height from the ground is the object no longer in contact with the sphere?
Answer:
10.2 m .
Explanation:
Let object falls by angle θ .
At any moment after the fall , there are two forces acting on the sphere
1 ) mg cosθ which is a component of weight towards the centre 2 ) normal reaction of the surface R .
mgcosθ - R is net force acting, which provides centripetal force
mgcosθ - R = mv² / r
But v² = 2g r( 1-cosθ ) [ object falls by height ( r - r cosθ ).
mgcosθ - R = m / r x 2g r( 1-cosθ )
When the object is no longer in touch with sphere,
R = 0
mgcosθ = m / r x 2g r( 1-cosθ )
3 gr cosθ = 2gr
cosθ = 2/3
height of fall
= r ( 1-cosθ )
r ( 1 - 2/3 )
1/3 r
1/3 x 15.3
5.1 m
Height from the ground
15.3 - 5.1
10.2 m .
The object sliding down the sphere loses contact at a point where its gravitational and centripetal forces become equal, corresponding to an angle θ where cosθ = 2/3. Given the sphere's radius of 15.3m, this corresponds to a height from the ground of 10.2 m.
Explanation:To solve this problem, we need to apply the concept of conservative forces and energy conservation. Initially, the object is at rest at the top of the sphere, thus it possesses gravitational potential energy and no kinetic energy. As the object slides down, its potential energy converts into kinetic energy until a point where the object no longer remains in contact with the sphere.
This occurs when the gravitational force acting towards the center of the sphere becomes equal to the centripetal force required for the object in circular motion. Mathematically, this correlates to the equation: mgcosθ = mv²/r.
By simplifying this equation, considering all the potential energy converts into kinetic energy at the point of leaving contact (mgh = 1/2mv²), and knowing the radius of the sphere, we can deduce that the object loses contact with the sphere at a point where the angle θ (from the vertical axis) is cosθ = 2/3. The height (h) it loses contact can then be determined by the relationship h = r(1 - cosθ). Given the radius r = 15.3m, we find the height from the ground to be 10.2 m.
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Earth turns on its axis about once every 24 hours. The Earth's equatorial radius is 6.38 x 106 m. If some catastrophe caused Earth to suddenly come to a screeching halt, with what speed would Earth's inhabitants who live at the equator go flying off Earth's surface?
To solve this problem it is necessary to apply the concepts related to the Period of a body and the relationship between angular velocity and linear velocity.
The angular velocity as a function of the period is described as
[tex]\omega = \frac{2\pi}{T}[/tex]
Where,
[tex]\omega =[/tex]Angular velocity
T = Period
At the same time the relationship between Angular velocity and linear velocity is described by the equation.
[tex]v = \omega r[/tex]
Where,
r = Radius
Our values are given as,
[tex]T = 24 hours[/tex]
[tex]T = 24hours (\frac{3600s}{1 hour})[/tex]
[tex]T = 86400s[/tex]
We also know that the radius of the earth (r) is approximately
[tex]6.38*10^6m[/tex]
Usando la ecuación de la velocidad angular entonces tenemos que
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{86400}[/tex]
[tex]\omega = 7.272*10^{-5}rad/s[/tex]
Then the linear velocity would be,
[tex]v = \omega *r[/tex]
x[tex]v = \omega *r[/tex]
[tex]v= 463.96m/s[/tex]
The speed would Earth's inhabitants who live at the equator go flying off Earth's surface is 463.96
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s in 0.820 s. (a) What is the magnitude of the linear impulse experienced by a 66.0-kg passenger in the car during the time the car accelerates?
Answer:
[tex]I=336.6kgm/s[/tex]
Explanation:
The equation for the linear impulse is as follows:
[tex]I=F\Delta t[/tex]
where [tex]I[/tex] is impulse, [tex]F[/tex] is the force, and [tex]\Delta t[/tex] is the change in time.
The force, according to Newton's second law:
[tex]F=ma[/tex]
and since [tex]a=\frac{v_{f}-v_{i}}{\Delta t}[/tex]
the force will be:
[tex]F=m(\frac{v_{f}-v_{i}}{\Delta t})[/tex]
replacing in the equation for impulse:
[tex]I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)[/tex]
we see that [tex]\Delta t[/tex] is canceled, so
[tex]I=m(v_{f}-v_{i})[/tex]
And according to the problem [tex]v_{i}=0m/s[/tex], [tex]v_{f}=5.10m/s[/tex] and the mass of the passenger is [tex]m=66kg[/tex]. Thus:
[tex]I=(66kg)(5.10m/s-0m/s)[/tex]
[tex]I=(66kg)(5.10m/s)[/tex]
[tex]I=336.6kgm/s[/tex]
the magnitude of the linear impulse experienced the passenger is [tex]336.6kgm/s[/tex]
Three cars (car L, car M, and car N) are moving with the same speed and slam on their brakes. The most massive car is car L, and the least massive is car N. If the tires of all three cars have identical coefficients of kinetic friction with the road surface, for which car is the amount of work done by friction in stopping it the greatest?A) The amount of work done by friction is the same for all cars.B) Car NC) Car MD) Car L
Answer:
I believe it's B
The amount of work done by friction in stopping a car is determined by the force of friction and the distance over which it acts. Car L will require the greatest force of friction to stop it, making the amount of work done by friction the greatest for Car L.
Explanation:The amount of work done by friction in stopping a car is determined by the force of friction and the distance over which it acts. Since all three cars have identical coefficients of kinetic friction with the road surface, the force of friction will be the same for all three cars. However, the stopping distance for each car will vary based on its mass.
Car L has the greatest mass, so it will require the greatest force of friction to stop it. Therefore, the amount of work done by friction in stopping car L will be the greatest among the three cars.
Car M and Car N will require less force of friction compared to car L, so the amount of work done by friction in stopping them will be less than that of Car L, but equal to each other.
The average density of the body of a fish is 1080 kg/m3. To keep from sinking, the fish increases its volumeby inflating an internal air bladder. By what percentage must the fish increase its volume in order to beneutrally buoyant in fresh water? The density of air at room temperature is 1.2 kg/m3.
Answer:
f = 8 %
Explanation:
given,
density of body of fish = 1080 kg/m³
density of air = 1.2 Kg/m³
density of water = 1000 kg/m²
to protect the fish from sinking volume should increased by the factor f
density of fish + density of water x increase factor = volume changes in water
1080 +f x 1.2 =(1 + f ) x 1000
1080 + f x 1.2 = 1000 + 1000 f
998.8 f = 80
f = 0.0800
f = 8 %
the volume increase factor of fish will be equal to f = 8 %
A space shuttle in orbit around the Earth carries its payload with its mechanical arm. Suddenly, the arm malfunctions and releases the payload. What will happen to the payload?
a. It will fall straight down and hit the Earth.
b. It will follow a curved path and eventually hit the Earth.
c. It will drift out into deep space.
d. It will remain in the same orbit with the shuttle.
We must remember that a body in Earth orbit is under perpetual free fall, that is, that the body moves under a constant speed in constant around the earth.
The moment one of the charges is released, its displacement speed will be equal to that of the ship, so it will tend to continue rotating in the orbit in which the ship was traveling.
The correct answer is D. The payload will remain in the same orbit with the space shuttle.
A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanbag is moving exactly horizontally when it strikes and sticks to the inside wall of the crate, which is lined with velcro. As a result of this impact, the crate slides backwards 0.935 meters along the concrete, which has a coefficient of kinetic friction of \mu_k = 0.480μ k =0.480. The beanbag has a mass of 0.354 kg and the empty crate has a mass of 3.77 kg. Calculate the speed of the beanbag vv at the moment it strikes the crate. You may neglect the effect of air resistance.
Answer:
31.42383 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
[tex]\mu[/tex] = Coefficient of kinetic friction = 0.48
s = Displacement = 0.935 m
[tex]m_1[/tex] = Mass of bean bag = 0.354 kg
[tex]m_2[/tex] = Mass of empty crate = 3.77 kg
[tex]v_1[/tex] = Speed of the bean bag
[tex]v_2[/tex] = Speed of the crate
Acceleration
[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]
[tex]a=--9.81\times 0.48=4.7088\ m/s^2[/tex]
From equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s[/tex]
In this system the momentum is conserved
[tex]m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s[/tex]
The speed of the bean bag is 31.42383 m/s
The speed of the beanbag when it strikes the crate is 34.46 m/s.
Inelastic collision:
In an inelastic collision, two bodies collide, stick together and move with the same velocity. The momentum is conserved while the kinetic energy is not.
As the beanbag sticks to the crate they move together with the same velocity until they stop after 0.935 m due to frictional force.
Let M be the combined mass of the crate and beanbag. the frictional force acting on the system is:
[tex]F=-\mu_k Mg\\\\Ma=-\mu_k Mg\\\\a=-\mu g\\\\a=-0.480\times9.8\;m/s^2\\\\a=4.7\;m/s^2[/tex]will be the acceleration of the system.
Let the initial velocity of the system be u and the final velocity is 0, then from the third equation of motion:
[tex]0=u^2-2as\\\\u=\sqrt[]{2as}\\\\u=\sqrt{2\times4.7\times0.935}\;m/s\\\\u=2.96\;m/s[/tex]
So the momentum after collision is
Mu = (0.354 + 3.77)×2.96 = 12.2 kgm/s
Which must be equal to the initial momentum that is contributed only by the beanbag, since the crate is on rest. Let the speed of the bean bag be v then initial momentum should be:
12.2 = 0.354 × v
v = 34.46 m/s
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A 0.750-kg object hanging from a vertical spring is observed to oscillate with a period of 1.50 s. When the 0.750-kg object is removed and replaced by a 1.50-kg object, what will be the period of oscillation
Answer:
New time period, [tex]T_2=2.12\ s[/tex]
Explanation:
Given that,
Mass of the object 1, [tex]m_1=0.75\ kg[/tex]
Time period, [tex]T_1=1.5\ s[/tex]
If object 1 is replaced by object 2, [tex]m_2=1.5\ kg[/tex]
Let [tex]T_2[/tex] is the new period of oscillation.
The time period of oscillation of mass 1 is given by :
[tex]T_1=2\pi \sqrt{\dfrac{m_1}{k}}[/tex]
[tex]1.5=2\pi \sqrt{\dfrac{0.75}{k}}[/tex]............(1)
The time period of oscillation of mass 2 is given by :
[tex]T_2=2\pi \sqrt{\dfrac{m_2}{k}}[/tex]
[tex]T_2=2\pi \sqrt{\dfrac{1.5}{k}}[/tex]............(2)
From equation (1) and (2) we get :
[tex](\dfrac{T_1}{T_2})^2=\dfrac{m_1}{m_2}[/tex]
[tex](\dfrac{1.5}{T_2})^2=\dfrac{0.75}{1.5}[/tex]
[tex]\dfrac{1.5}{T_2}=0.707[/tex]
[tex]T_2=2.12\ s[/tex]
So, the new period of oscillation is 2.12 seconds. Hence, this is the required solution.
Final answer:
To find the new period of oscillation when the mass attached to a spring is increased, one can use the formula for the period T = 2π√(m/k). When the mass is doubled from 0.750 kg to 1.50 kg, the new period of oscillation is approximately 2.12 seconds.
Explanation:
The period of oscillation for a mass m suspended from a spring is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Using this relationship, we can compare the periods for different masses to find the new period when the mass changes.
When the mass was 0.750 kg, the period was 1.50 s. Now, if the mass is increased to 1.50 kg, we can expect the period to change. Noting that the spring constant k remains the same, we use the ratio of the periods T1/T2 = √(m1/m2). Solving for the new period T2 gives us T2 = T1 * √(m2/m1).
If we plug in the numbers, we get:
T1 = 1.50 s (given for 0.750 kg)
m1 = 0.750 kg
m2 = 1.50 kg
T2 = 1.50 s * √(1.50 kg / 0.750 kg) = 1.50 s * √2 ≈ 2.12 s
The new period of oscillation for the 1.50-kg object is therefore approximately 2.12 seconds.
On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outside the window moves with a speed of approximately 150m/s shortly after takeoff, and that the air inside the plane is at atmospheric pressure.(A) Find the pressure difference between the inside and outside of the window.(B) If the window is 25cm by 45cm, find the force exerted on the window by air pressure.
The pressure difference between the inside and outside of the window can be found using Bernoulli's principle. The force exerted on the window by air pressure can be calculated using the formula F = P * A.
Explanation:A) The pressure difference between the inside and outside of the window can be found using Bernoulli's principle. According to Bernoulli's principle, the pressure difference is equal to the difference in kinetic energy per unit volume between the inside and outside air. This can be expressed as: ΔP = 1/2 ρv², where ΔP is the pressure difference, ρ is the density of air, and v is the speed of air. Substituting the given values, we get: ΔP = 1/2 * 1.14 kg/m³ * (150 m/s)² = 12,825 Pa.
B) The force exerted on the window by air pressure can be calculated using the formula: F = P * A, where F is the force, P is the pressure difference, and A is the area of the window. Substituting the given values, we get: F = 12,825 Pa * (0.25 m * 0.45 m) = 1,826.25 N.
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The driver of a car, which is initially at rest at the top A of the grade, releases the brakes and coasts down the grade with an acceleration in feet per second squared given by a = 3.22-0.004v2 , where v is the velocity in feet per second. Determine the velocity vB at the bottom B of the grade.
Answer:
28.3 ft/s
Explanation:
We are given that
Initial velocity of a car=[tex]u=0[/tex]
Acceleration of the car=[tex]a=3.22-0.004v^2[/tex]
We have to find the velocity [tex]v_B[/tex] at the bottom of the grade.
Distance covered by car=600 ft
We know that
Acceleration=a=[tex]\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}[/tex]
[tex]v=\frac{ds}{dt}[/tex]
[tex]ds=\frac{vdv}{a}[/tex]
Taking integration on both side and taking limit of s from 0 to 600 and limit of v from 0 to [tex]v_B[/tex]
[tex]\int_{0}^{600}ds=\int_{0}^{v_B}\frac{vdv}{3.22-0.004v^2}[/tex]
[tex][S]^{600}_{0}=-\frac{1}{0.008}[ln(3.22-0.004v^2)]^{v_B}_{0}[/tex]
Using substitution method and [tex]\int f(x)dx=F(b)-F(a)[/tex]
[tex]600-0=-\frac{1}{0.008}[ln(3.22-0.004v^2_B)-ln(3.22)][/tex]
[tex]600=-\frac{1}{0.008}(ln(\frac{3.22-0.04v_B}{3.22}))[/tex]
[tex]ln(m)-ln(n)=ln(\frac{m}{n})[/tex]
[tex]600\times 0.008=-ln(\frac{3.22-0.04v_B}{3.22})[/tex]
[tex]-4.8=ln(\frac{3.22-0.04v_B}{3.22})[/tex]
[tex]\frac{3.22-0.04v_B}{3.22}=e^{-4.8}[/tex]
[tex]ln x=y\implies x=e^y[/tex]
[tex]3.22-0.004v^2_B=3.22e^{-4.8}[/tex]
[tex]3.22-0.004v^2_B=0.0265[/tex]
[tex]3.22-0.0265=0.004v^2_B[/tex]
[tex]0.004v^2_B=3.22-0.0265=3.1935[/tex]
[tex]v_B=\sqrt{\frac{3.1935}{0.004}}=28.3 ft/s[/tex]
Hence, the velocity [tex]v_B[/tex] at the bottom of the grade=28.3 ft/s
The velocity of the car at the bottom of the grade is 28.37 ft/s.
What is acceleration?Acceleration is the rate of change of velocity with time.
Motion of the car from top to bottom of the gradeThe velocity of the car increases as the car moves downwards and it will be maximum when the car reaches equilibrium position. At equilibrium position (bottom of the grade) the acceleration will be zero but the velocity will be maximum.
a = 3.22 - 0.004v²
0 = 3.22 - 0.004v²
0.004v² = 3.22
v² = 3.22/0.004
v² = 805
v = √805
v = 28.37 ft/s
Thus, the velocity of the car at the bottom of the grade is 28.37 ft/s.
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A 1-kg ball is released from a height of 6 m, and a 2-kg ball is released from a height of 3 m. Air resistance is negligible as they fall. Which of the following statements about these balls are correct?
A) Both balls will reach the ground with the same kinetic energy.
B) Both balls will reach the ground with the same speed.
C) As they reach the ground, the 1-kg ball will be moving faster than the 2-kg ball.
D) As they reach the ground, the 1-kg ball will have more kinetic energy than the 2-kg ball because it was dropped from a greater height.
E) Both balls will take the same time to reach the ground.
Answer:
A) True B) false C) True D) False E) False
Explanation:
A) We know that the potential energy is transformed into kinetic energy since the ball is released.
[tex]EP1=m*g*h = 1 * 9.81 * 6 = 58.86J\\[/tex]
This energy will be the same at the moment that the ball is about to hit the floor. Therefore there was a transformation from potential to kinetic energy
EK1 = EP1
[tex]EK=\frac{1}{2}*m*v^{2} \\v1=\sqrt{(58.86*2)/1} \\v1=10,84 m/s[/tex]
With the 2kg ball happens the same
[tex]EP2=m*g*h2=2*9.81*3=58.86J\\[/tex]
Velocity at the moment when is about to hit the floor (due to the energy transform) will be:
[tex]EK2=\frac{1}{2}*m*v^{2} \\v1=\sqrt{(58.86*2)/2} \\v2=7.67 m/s[/tex]
A) Since both balls have the same potential energy, they will have the same kinetic energy
B) The velocities are differents as they were calculated before.
C) Therefore the 1 kg ball moves faster as it reaches the ground.
D) The kinetic energy is the same for both balls as it was calculated before.
E) The 2 kg ball will reach the ground first, since it is closer to the ground
With the velocities calculated in the previous steps we have now:
[tex]1 kg ball\\v=v0+g*t\\t1=10.85/9.81 = 1.1s\\\\2 kg ball\\t2=7.672/9.81 = 0.782s[/tex]
A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 170 N at an angle of 25◦ above the horizontal. The box has a mass of 27.4 kg, and the coefficient of friction between box and floor is 0.293. The acceleration of gravity is 9.8 m/s 2 . 27.4 kg µ = 0.293 170 N 25◦ Find the acceleration of the box. Answer in units of m/s 2 .
Answer:
a = 3.52 m/s²
Explanation:
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
m= 27.4 kg : mas of the box
F= 170 N, at an angle of 25◦ above the horizontal :Force rope attached to the box
μk = 0.293 :Coefficient of friction between box and floor
g = 9.8 m/s² : acceleration due to gravity
We define the x-axis in the direction parallel to the movement of the box and the y-axis in the direction perpendicular to it.
Forces acting on the box
W: Weight of the block : In vertical direction ,downward
FN : Normal force : perpendicular to the floor upward
f : Friction force: parallel to the floor and opposite to the movement
F : force of the rope attached to the box , at an angle of 25◦ above the horizontal
Calculated of the W ( weight of the box)
W= m*g
W= 27.4 kg* 9.8 m/s² = 268.52 N
x-y components of the force of 170 N
Fx=170 N *cos 25° = 154.07 N
Fy=170 N *sin 25° =71.845 N
Calculated of the FN ( Normal force)
We apply the formula (1)
∑Fy = m*ay ay = 0
FN + Fy - W = 0
FN = W- Fy
FN = 268.52 N - 71.845 N
FN =196. 675 N
Calculated of the f (friction force)
f = μk*FN
f = 0.293*196. 675
f = 57.626 N
We apply the formula (1) to calculated acceleration of the box:
∑Fx = m*ax , ax= a : acceleration of the box
Fx-f = m*a
154.07-57.626 = (27.4)*a
96.45 = (27.4)*a
a = (96.45)/ (27.4)
a = 3.52 m/s²
Final answer:
To calculate the box's acceleration, resolve the pulling force into components, find the normal and frictional forces, and apply Newton's second law with the net horizontal force and box's mass.
Explanation:
To find the acceleration of the box, we first need to resolve the applied force into horizontal and vertical components. The horizontal component of the force ( extit{F}_{horizontal}) is calculated by 170 N imes extit{cos}(25°), and the vertical component ( extit{F}_{vertical}) by 170 N imes extit{sin}(25°). The normal force ( extit{N}) is the sum of the vertical component and the weight of the box acting upwards, which is the box mass (27.4 kg) multiplied by the acceleration of gravity (9.8 m/s²), but since the box doesn't accelerate vertically, the normal force is equal to the weight minus extit{F}_{vertical}.
Next, we calculate the frictional force ( extit{F}_{friction}) using the coefficient of friction (0.293) and the normal force. The net horizontal force ( extit{F}_{net}) is the difference between the horizontal component of the pull and the frictional force. Finally, we use Newton's second law ( extit{F} = m imes extit{a}) to find the acceleration by dividing the extit{F}_{net} by the mass of the box (27.4 kg).
A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2l. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.789 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.
Answer:
Explanation:
For rate of flow of liquid in a pipe , the formula is
Q = π p r⁴ / 8 η l
where Q is volume of liquid flowing per unit time , p is pressure drop across pipe , r is radius of pipe , η is coefficient of viscosity and l is length of pipe
If η and l are constant for two pipes having equal rate of flow as in the given case
p₁ r₁⁴ = p₂ r₂⁴
p₁ = 1.789 p₂ ( given )
r₁ = D/2
1.789 p₂ x (D/2)⁴ = p₂ x r₂⁴
r₂ = [tex]\frac{D}{2}\times\sqrt[4 ]{1.789}[/tex]
2 x r₂ = D x 1.1565
D₂ = D x 1.1565
To determine the diameter of the second pipe, apply Poiseuille's law to relate the known pressure drop ratio to the radii of the pipes, and then solve for the radius, r², of the second pipe.
Explanation:The subject of this question pertains to the Physics of fluid mechanics, specifically dealing with the laminar flow of a fluid through pipes of different diameters and the associated pressure drops.
According to Poiseuille's law, which describes laminar flow in pipes, the pressure drop (ΔP) across a length of pipe is directly proportional to the length (L) and the viscosity (η) of the fluid, and inversely proportional to the fourth power of the radius (r) of the pipe.
Given that the pressure drop for the first pipe with diameter D is 1.789 times greater than the second pipe, we can establish the following relationship using Poiseuille's equation ΔP ∝ Lη/r^4 (ignoring constants for convenience since lengths are equal, and assuming the same fluid viscosity):
(ΔP1/ΔP2) = (r2⁴ / r1⁴)
1.789 = (r2⁴ / (D/2)⁴)
Therefore, solving for r2 (radius of the second pipe), we get:
r²= 0.5 * D * (1.789)^(1/4)
Let d2 be the diameter of the second pipe, then d2 = 2 * r2:
d2 = D * (1.789)^(1/4)
By calculating the fourth root of 1.789 and multiplying it by the diameter D of the first pipe, the diameter of the second pipe can be determined.
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When you take your 1900-kg car out for a spin, you go around a corner of radius 53m with a speed of 13m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesnt skid, what is the force exerted on it by static friction?
Answer:
Ff = 6058.5N
Explanation:
The sum of forces is:
[tex]Ff = m*a_c[/tex]
[tex]Ff = m*V^2/R[/tex]
[tex]Ff = 1900*13^2/53[/tex]
[tex]Ff = 6058.5N[/tex]
A construction worker uses a steel tape to measure the length of an aluminum support column. If the measured length is 17.700 m when the temperature is 21.2°C, what is the measured length when the temperature rises to 29.4°C? (Note: Don't neglect the expansion of the steel tape. Give your answer to three decimal places.)
The measured length when the temperature rises to 29.4°C is 17.712 m.
Explanation:To answer this question, we need to consider the thermal expansion of the steel tape and the aluminum support column. The coefficient of linear expansion for the steel tape is not given, so we will assume it to be the same as iron, which is approximately 12x10-6/°C.
We can use the formula:
ΔL = αL₀ΔT
where ΔL is the change in length, α is the coefficient of linear expansion, L₀ is the initial length, and ΔT is the change in temperature.
Given that the initial length (L₀) is 17.700 m and the initial temperature is 21.2°C, we can calculate the change in length (ΔL) when the temperature rises to 29.4°C.
ΔL = (12x10-6)(17.700)(29.4 - 21.2) = 0.01226352 m
Therefore, the measured length when the temperature rises to 29.4°C is 17.700 + 0.01226352 = 17.712 m (rounded to three decimal places).
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The change in length of the aluminum column when the temperature rises can be found using the equation ∆L = αL0∆T, where ∆L is the change in length, α is the coefficient of linear expansion, L0 is the initial length, and ∆T is the change in temperature. Plugging in the values provided, we find the change in length is 3.25848 x 10-4 m, rounded to three decimal places.
Explanation:When a material undergoes thermal expansion, its length increases as the temperature rises. To calculate the change in length, we can use the equation ∆L = αL0∆T, where ∆L is the change in length, α is the coefficient of linear expansion, L0 is the initial length, and ∆T is the change in temperature. In this case, we need to find the change in length of the aluminum column when the temperature rises from 21.2°C to 29.4°C.
First, we need to determine the coefficient of linear expansion for aluminum. Using the information provided, we can find that the coefficient of linear expansion for aluminum is approximately 22·10-6 °C-1.
Next, we can plug in the values into the equation. Given L0 = 17.700 m, ∆T = (29.4 - 21.2)°C = 8.2°C, and α = 22·10-6 °C-1, we can calculate ∆L as follows:
∆L = (22·10-6 °C-1)(17.700 m)(8.2°C) = 3.25848 x 10-4 m
Rounded to three decimal places, the measured length of the aluminum support column when the temperature rises to 29.4°C is 0.000 m.
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In a double-star system, two stars of mass 4.6 x 1030 kg each rotate about the system's center of mass at radius 1.9 x 1011 m.
(a) What is their common angular speed?
(b) If a meteoroid passes through the system's center of mass perpendicular to their orbital plane, what minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system?
To carry out this exercise, it is necessary to use the equations made to Centripetal Force and Gravitational Energy Conservation.
By definition we know that the Centripetal Force is estimated as
[tex]F_c = M\omega^2R[/tex]
Where,
M = mass
[tex]\omega =[/tex] Angular velocity
R = Radius
From the 'linear' point of view the centripetal force can also be defined as
[tex]F_c = \frac{GM^2}{R^2}[/tex]
PART A ) Equating both equations we have,
[tex]\frac{M}{\omega^2R}=\frac{GM^2}{R^2}[/tex]
Re-arrange to find \omega
[tex]\omega = \sqrt{\frac{Gm}{r^3}}[/tex]
Replacing with our values
[tex]\omega = \sqrt{\frac{(6.67*10^{-11})(4.6*10^{30})}{(1.9*10^{11})^3}}[/tex]
[tex]\omega = 2.115*!0^{-7}rad/s[/tex]
Therefore the angular speed is [tex]\omega = 2.115*!0^{-7}rad/s[/tex]
PART B) For energy conservation we have to
[tex]KE_{min} = PE_{cm}[/tex]
Where,
[tex]KE_{min} =[/tex] Minimus Kinetic Energy
[tex]PE_{cm} =[/tex] Gravitational potential energy at the center of mass
Then,
[tex]\frac{1}{2} mv^2_{min} = \frac{2GMm}{R}[/tex]
Re-arrange to find v,
[tex]v_min = \sqrt{\frac{4GM}{R}}[/tex]
[tex]v_min = \sqrt{\frac{4(6.67*10^{-11})(2.2*10^{30})}{(1.9*10^{11})}}[/tex]
[tex]v_min = 5.55*10^4m/s[/tex]
Therefore the minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system is [tex]v_min = 5.55*10^4m/s[/tex]
In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.
Answer:
Explanation:
Let length of the pendulum be l . The expression for time period of pendulum is as follows
T = 2π[tex]\sqrt{\frac{l}{g} }[/tex]
For Mars planet ,
1.5 = [tex]2\pi\sqrt{\frac{l}{.38\times9.8} }[/tex]
For other planet
.92 = [tex]2\pi\sqrt{\frac{l}{g_1} }[/tex]
Squiring and dividing the two equations
[tex]\frac{1.5^2}{.92^2} = \frac{g_1}{3.8\times9.8}[/tex]
[tex]g_1 = 9.9[/tex]
The second planet appears to be earth.
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at rest. After he lands on the raft, the raft, with him on it, moves away from the pier at 1.6 m/s. What was Tyrone's speed as he ran off the end of the pier?
Answer:u=4.04 m/s
Explanation:
Given
Mass m=85 kg
mass of Raft M=130 kg
velocity of raft and man v=1.6 m/s
Let initial speed of Tyrone is u
Conserving Momentum as there is no external Force
[tex]mu=(M+m)v[/tex]
[tex]85\times u=(85+130)\cdot 1.6[/tex]
[tex]u=\frac{215}{85}\cdot 1.6[/tex]
[tex]u=2.529\cdot 1.6=4.04 m/s[/tex]
Final answer:
Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.
Explanation:
To calculate Tyrone's speed as he ran off the end of the pier, we can use the principle of conservation of momentum. According to this principle, the initial momentum of the system (Tyrone and the raft) is equal to the final momentum. Tyrone's initial momentum is given by his mass (85 kg) multiplied by his speed (which we need to find). The mass of the raft is 130 kg and its initial velocity is 0 m/s since it was at rest. The final momentum of the system is given by the mass of Tyrone and the raft (215 kg) multiplied by their final velocity (1.6 m/s).
Using the equation for conservation of momentum, we have:
(85 kg)(v) = (215 kg)(1.6 m/s)
Solving for the velocity v:
v = (215 kg)(1.6 m/s) / 85 kg
v ≈ 4.04 m/s
Therefore, Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.