Answer:
Final velocity at the bottom of hill is 15.56 m/s.
Explanation:
The given problem can be divided into four parts:
1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)
From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:
[tex]p_i = p_f[/tex]
[tex]m_1u_1 + m_2v_2 = (m_1 + m_2)v[/tex]
[tex]v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)} [/tex]
[tex]v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s[/tex]
2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.
The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:
[tex]E(i) = E(f)[/tex]
[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]
[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]
[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}[/tex]
Here, initial velocity is the final velocity from the first stage. Therefore:
[tex]v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s[/tex]
3. Use conservation of momentum to find the combined speed of Gayle and her brother.
Given:
Initial velocity of Gayle and sled is, [tex]u_1(i)=10.5[/tex] m/s
Initial velocity of her brother is, [tex]u_2(i)=0[/tex] m/s
Mass of Gayle and sled is, [tex]m_1=55.0[/tex] kg
Mass of her brother is, [tex]m_2=30.0[/tex] kg
Final combined velocity is given as:
[tex]v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}[/tex]
[tex]v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 [/tex] m/s
4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.
Using conservation of energy, the final velocity at the bottom of the hill is:
[tex]E(i) = E(f)[/tex]
[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]
[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]
[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s[/tex]
Using principles of kinetic and potential energy, and assuming ideal conditions devoid of friction and air resistance, Gayle and her brother's final speed would be equal to the square root of twice the product of acceleration due to gravity and the total height of the hill.
Explanation:This question relates to the principles of kinetic energy and potential energy, as well as conservation of energy. When Gayle initially sleds down the hill, she converts potential energy (mgh) into kinetic energy (1/2 mv^2). Taking 'g' as acceleration due to gravity (9.8 m/s^2) and 'h' as the vertical height she descends (5.0 m), the speed she attains at the time her brother joins (v') can be calculated using √(2gh).
The potential energy at that point is their combined mass times gravity and their height, which is transformed to kinetic energy at the bottom. If 'H' is the total height of the hill (15.0 m), their final combined speed is given by √(2gH). Ignoring resistive forces, they maintain this speed for the entire ride down the hill.
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You push very hard on a heavy desk, trying to move it. You do work on the desk:
a. only if it doesn't move.
b. whether or not it moves, as long as you are exerting a force.
c. only if it starts moving.
d. never—it does work on you.
Answer:
(C) Only if it starts moving
Explanation:
We know that work done is given by
[tex]W=F.d=Fdcos\Theta[/tex]
So there are two case in which work done is zero
First case is that when force and displacement are perpendicular to each other
And other case is that when there is no displacement
So for work to be done there must have displacement, if there is no displacement then there is no work done
So option (c) will be the correct option
Work in a physical sense is done on a desk only if the desk moves when you push it. Simply exerting force is not considered work without displacement in the direction of the force. So the correct option is C.
Explanation:When considering whether you do work on a heavy desk by pushing it, it's important to understand the scientific definition of work. In physics, work is defined as the transfer of energy that occurs when a force is applied over a distance. Therefore, you do work on an object if, and only if, the object moves in the direction of the force. So the correct answer to the question is:
c. only if it starts moving.
Simply exerting a force on an object does not constitute work unless the object is displaced. For example, if you continue to push against a wall and it does not move, despite your effort and energy consumption, physically no work is done because there is no displacement.
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For a particular reaction, Δ H ∘ = − 93.8 kJ and Δ S ∘ = − 156.1 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?
To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity
At constant temperature and pressure, the change in Gibbs free energy is defined as
[tex]\Delta G = \Delta H - T\Delta S[/tex]
Where,
H = Entalpy
T = Temperature
S = Entropy
When the temperature is less than that number it is negative meaning it is a spontaneous reaction. [tex]\Delta G[/tex] is also always 0 when using single element reactions. In numerical that implies [tex]\Delta G = 0[/tex]
At the equation then,
[tex]\Delta G = \Delta H - T\Delta S[/tex]
[tex]0 = \Delta H - T\Delta S[/tex]
[tex]\Delta H = T\Delta S[/tex]
[tex]T = \frac{\Delta H}{\Delta S}[/tex]
[tex]T = \frac{-93.8kJ}{-156.1J/K}[/tex]
[tex]T = \frac{-93.8*10^3J}{-156.1J/K}[/tex]
[tex]T = 600.89K[/tex]}
Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
What is the relation between the standard free Gibbs energy and the spontaneity of a reaction?When ΔG° < 0, the reaction is spontaneous.When ΔG° = 0, the reaction is not spontaneous nor nonspontaneous.When ΔG° > 0, the reaction is nonspontaneous.We want to know at what temperature a reaction changes from nonspontaneous to spontaneous, that is, at what temperature ΔG° = 0.
Given the standard enthalpy and entropy of the reaction, we can calculate that temperature using the following expression.
ΔG° = ΔH° - T . ΔS°
0 = -93.8 kJ - T . (-156.1 J/K)
T = 601 K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
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The production of sound during speech or singing is a complicated process. Let's concentrate on the mouth. A typical depth for the human mouth is about 8.00 cm, although this number can vary. (Check it against your own mouth.) We can model the mouth as an organ pipe that is open at the back of the throat. A) What are the wavelengths of the first four harmonics you can produce if your mouth is open? (in cm/4 answers) B)What are the frequencies of the first four harmonics you can produce if your mouth is open? Use 354 (in Hz/4 answers) C)What are the wavelengths of the first four harmonics you can produce if your mouth is closed? (in cm/4 answers) D)What are the frequencies of the first four harmonics you can produce if your mouth is closed?
Answer:A) The wavelengths are:8cm for first harmonic
5.3cm for second harmonic
4cm for third harmonic
3.2cm for fourth harmonic
B) The frequencies are:
44.25Hz for first harmonic
66.38Hz for second harmonic
88.5Hz for third harmonic
110.63Hz for fourth harmonic
C) The wavelengths are:
10.66Cm for first harmonic
6.4 cm for second harmonic
4.57cm for third harmonic
3.5cm for fourth harmonic
D) The frequencies are:
33.19Hz for first harmonic
55.31 Hz for second harmonic
77.44Hz for third harmonic
99.56Hz for fourth harmonic
Explanation: since the mouth is being modelled as an organ pipe, then we can say that an open mouth is a model of an organ pipe open at both ends.
For an open pipe all harmonics are possible but only odd harmonic's are possible in closed pipes. A closed mouth is a model of a closed pipe, since closed pipes are open at one end.
In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the spring of 9.50 cm is recorded. With 1.00 kg hanging on the spring, a second elongation of 12.00 cm is recorded. Calculate the spring constant k in Newtons per meter (N/m). (Note: The equilibrium position is not zero.)
To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.
By definition we know that Newton's second law is defined as
[tex]F = ma[/tex]
m = mass
a = Acceleration
By Hooke's law force is described as
[tex]F = k\Delta x[/tex]
Here,
k = Gravitational constant
x = Displacement
To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.
The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so
[tex]k\Delta x = mg[/tex]
So for state 1 we have that with 0.2kg there is an elongation of 9.5cm
[tex]k (9.5-l)=0.2*g[/tex]
[tex]k (9.5-l)=0.2*9.8[/tex]
For state 2 we have that with 1Kg there is an elongation of 12cm
[tex]k (12-l)= 1*g[/tex]
[tex]k (12-l)= 1*9.8[/tex]
We have two equations with two unknowns therefore solving for both,
[tex]k = 3.136N/cm[/tex]
[tex]l = 8.877cm[/tex]
In this way converting the units,
[tex]k = 3.136N/cm(\frac{100cm}{1m})[/tex]
[tex]k = 313.6N/m[/tex]
Therefore the spring constant is 313.6N/m
A ring made from brass has an inner radius of 2.50000 cm and an outer radius of 3.50000 cm, giving the ring a thickness of 1.00000 cm. The thermal expansion coefficient of brass is 0.000019/°C. If the temperature of the ring is increased from 20.0°C to 90.0°C, by how much does the thickness of the ring change? _______ cm
Answer:0.00133 cm
Explanation:
Given
change in temperature [tex]\Delta T=90-20=70^{\circ}C[/tex]
Thermal Expansion coefficient of brass [tex]\alpha =0.000019 /^{\circ}C[/tex]
thickness t=1 cm
change in thickness is given by
[tex]\Delta t=t\cdot \alpha \cdot \Delta T[/tex]
[tex]\Delta t=1\times 0.000019\times 70[/tex]
[tex]\Delta t=0.00133\ cm[/tex]
A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effective value of 110 V is applied to the primary coil. At the secondary coil, an alternating voltage with an effective value of 550 V is obtained. A current of 29 A is supplied to the primary coil of the transformer. Calculate the maximum effective current in the secondary coil. The maximum effective current in the secondary coil is A.
Answer:
[tex]I_s=5.8A[/tex]
Explanation:
Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:
[tex]P_p=P_s[/tex]
So:
[tex]V_p*I_p=V_s*I_s[/tex]
Where:
[tex]V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil[/tex]
Solving for [tex]I_s[/tex]
[tex]I_s=\frac{V_p*I_p}{V_s}[/tex]
Replacing the data provided:
[tex]I_s=\frac{110*29}{550} =5.8A[/tex]
19.A 20 kg sign is pulled by a horizontal force such that the single rope (originally vertical) holding the sign makes an angle of 21° with the vertical. Assuming the sign is motionless, findA) the magnitude of the tension in the rope andB) the magnitude of the horizontal force.
Answer:
A)[tex]T=209.94N[/tex]
B) [tex]F=75.24N[/tex]
Explanation:
Using the free body diagram and according to Newton's first law, we have:
[tex]\sum F_y=Tcos(21^\circ)-mg=0(1)\\\sum F_x=F-Tsin(21^\circ)=0(2)[/tex]
A) Solving (1) for T:
[tex]T=\frac{mg}{cos(21^\circ)}\\T=\frac{20kg(9.8\frac{m}{s^2})}{cos(21^\circ)}\\T=209.94N[/tex]
B) Solving (2) for F:
[tex]F=Tsin(21^\circ)\\F=(209.94N)sin(21^\circ)\\F=75.24N[/tex]
A machine carries a 33.6 kg package from an initial position of r0 = (0.502 + 0.751 + 0.207) m at t0 = 0 s to a final position of r1 = (7.82 + 2.17 + 7.44) m at t1 = 11.9 s. The constant force applied by the machine on the package is F = (21.5 + 42.5 + 63.5) N.
Answer:
The work done on the package by the machine's force is 676.94 J.
Explanation:
Given that,
Mass of package = 33.6 kg
Initial position [tex]r_{0}=(0.502i+0.751j+0.207k)\ m[/tex]
Final position [tex]r_{1}=(7.82i+2.17j+7.44k)\ m[/tex]
Final time = 11.9 s
Force [tex]F=(21.5i+42.5j+63.5k)[/tex]
Suppose we need to find the work done on the package by the machine's force
We need to calculate the displacement
Using formula of displacement
[tex]d=r_{1}-r_{0}[/tex]
Put the value into the formula
[tex]d=(7.82i+2.17j+7.44k)-(0.502i+0.751j+0.207k)[/tex]
[tex]d=7.318i+1.419j+7.233k[/tex]
We need to calculate the work done
Using formula of work done
[tex]W=F\dotc d[/tex]
[tex]W=(21.5i+42.5j+63.5k)\dotc(7.318i+1.419j+7.233k)[/tex]
[tex]W=676.94\ J[/tex]
Hence, The work done on the package by the machine's force is 676.94 J.
A window in a house has a rectangular shape of 2.0 m by 1.0 m. The glass in the window is 0.5 cm thick, with a thermal conductivity of 0.84 W/(m oC). The temperature in the house is 20.5 °C, and the temperature outside is 4 oC. How much heat is lost through this window in one minute? Don't use scientific notations
Answer:
P = 5280 W
Explanation:
The conductivity of the materials determines that heat flows from the hot part to the cold part, the equation for thermal conductivity transfer is
P = Q / t = k A ([tex]T_{h}[/tex] -[tex]T_{c}[/tex]) / L
Where k is the thermal conductivity of the glass 0.8 W / ºC, A the area of the window, T the temperature and L is glass thickness
Let's calculate the window area
A = l * a
A = 2.0 1.0
A = 2.0 m²
Let's replace
L = 0.5 cm (1 m / 100 cm) = 0.005 m
P = 0.8 2 (20.5 - 4) / 0.005
P = 5280 W
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6640 N (1493 lbf). If the length of the rod is 370 mm (14.57 in.), what must be the diameter to allow an elongation of 0.53 mm (0.02087 in.)?
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
[tex]\Delta L=\dfrac{PL}{AE}[/tex]
[tex]A=\dfrac{PL}{\Delta LE}[/tex]
[tex]A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}[/tex]
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
Answer:
[tex]d=7.32\ mm[/tex]
Explanation:
Given:
Young's modulus, [tex]E=110\times 10^3\ MPa[/tex]yield strength, [tex]\sigma_y=240\ MPa[/tex]load applied, [tex]F=6640\ N[/tex]initial length of rod, [tex]l=370\ mm[/tex]elongation allowed, [tex]\Delta l=0.53[/tex]We know,
Stress:
[tex]\sigma=\frac{F}{A}[/tex]
where: A = cross sectional area
Strain:
[tex]\epsilon = \frac{\Delta l}{l}[/tex]
& by Hooke's Law within the elastic limits:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]\therefore 110\times 10^3=\frac{F}{A}\div \frac{\Delta l}{l}[/tex]
[tex]\therefore 110\times 10^3=\frac{6640\times 4}{\pi.d^2}\div \frac{0.53}{370}[/tex]
where: d = diameter of the copper rod
[tex]d=7.32\ mm[/tex]
A frictionless piston-cylinder device contains 4.5 kg of nitrogen at 110 kPa and 200 K. Nitrogen is now compressed slowly according to the relation PV1.5 = constant until it reaches a final temperature of 360 K. Calculate the work input during the process, in kJ.
Answer:
427.392 kJ
Explanation:
m = Mass of gas = 4.5 kg
Initial temperature = 200 K
Final temperature = 360 K
R = Mass specific gas constant = 296.8 J/kgK
[tex]\gamma[/tex] = Specific heat ratio = 1.5
Work done for a polytropic process is given by
[tex]W=\frac{mR\Delta T}{1-\gamma}\\\Rightarrow W=\frac{4.5\times 296.8(360-200)}{1-1.5}\\\Rightarrow W=-427392\ J\\\Rightarrow W=-427.392\ kJ[/tex]
The work input during the process is -427.392 kJ
Two castings are proposed for a study. One casting is a cube 10.16 cm (4 in.) on a side; the other is a rectangle 10.64 cm (16 in.) tall × 10.16 cm (4 in.) wide × 2.45 cm (1 in.) thick. Both are cast from the same metal at the same temperature, and the mold material is the same in both cases. If the cube-shaped casting solidifies in 14.5 min, how rapidly do we expect the rectangular casting to solidify?
To solve the problem it is necessary to apply the concepts related to Chvorinov's Law, which states that
[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]
Where,
[tex]V_c[/tex] = Volume cube
[tex]SA_c[/tex] = Superficial Area from Cube
[tex]V_r[/tex] = Volume Rectangle
[tex]SA_r[/tex]= Superficial Area from Rectangle
Our values are given as (I will try to develop the problem in English units for ease of calculations),
[tex]V_c = 4^3 = 64in^3[/tex]
[tex]SA_c = 6*4^2=96in^2[/tex]
[tex]SA_r = 2*(1*16+4*1+16*4)=168in^2[/tex]
[tex]V_r = 4*1*16 = 64in^3[/tex]
Applying the Chvorinov equation we have to,
[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]
[tex]\frac{T_r}{T_c} = (\frac{64}{64}*\frac{96}{168})^2[/tex]
[tex]\frac{T_r}{T_c} = (\frac{96}{168})^2[/tex]
[tex]\frac{T_r}{T_c} = 0.3265[/tex]
The stipulated time for the cube is 14.5 then,
[tex]T_r = 0.3265*14.5[/tex]
[tex]T_r = 4.735min[/tex]
What is the pressure (in N/m2) inside an alveolus having a radius of 2.22 ✕ 10−4 m if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as created by a spherical bubble.
Answer:
The pressure inside the bubble is 666.67 [tex]N/m^{2}[/tex]
Solution:
As per the question:
Radius, R = [tex]2.22\times 10^{- 4}\ m[/tex]
Now,
Given that the surface tension of the wall is the same as that of soapy water.
The air trapped inside the bubble exerts pressure on the soap bubble which is given by:
Gauge Pressure, P = [tex]\frac{4T}{r}[/tex]
Also, the surface tension of the soapy water, [tex]T_{s} = 0.0370\ N/m^{2}[/tex]
To calculate the pressure inside the alveolus:
[tex]P_{i} = \frac{4T}{R} = \frac{4T_{s}}{R}[/tex]
[tex]P_{i} = \frac{4\times 0.0370}{2.22\times 10^{- 4}} = 666.67\ N/m^{2}[/tex]
Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a θ = 58.4° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver?
Answer:
U = -3978.8 J
Explanation:
The work of the gravitational force U just depends of the heigth and is calculated as:
U = -mgh
Where m is the mass, g is the gravitational acceleration and h the alture.
for calculate the alture we will use the following equation:
h = L-Lcos(θ)
Where L is the large of the rope and θ is the angle.
Replacing data:
h = 12.2-12.2cos(58.4)
h = 5.8 m
Finally U is equal to:
U = -70(9.8)(5.8)
U = -3,978.8 J
Calculate the acceleration of gravity on the surface of the Sun. The mass of the Sun is MSun = 1.99 ✕ 1030 kg, the radius of the Sun is rSun = 6.96 ✕ 108 m, and G = 6.67 ✕ 10−11 N · m2/kg2. m/s2 (b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you can't.) Fg, Sun Fg, Earth =
Answer: 28
Explanation:
According to Newton's law of Gravitation the gravitational force on the surface of the Sun [tex]Fg_{s}[/tex] is:
[tex]Fg_{s}=mg_{sun}=G\frac{Mm}{r^{2}}[/tex] (1)
Where:
[tex]m[/tex] is your mass
[tex]g_{sun}[/tex] is the acceleration due gravity on the surface of the Sun
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the gravitational constant
[tex]M=1.99(10)^{30} kg[/tex] is the mass of the Sun
[tex]r=6.96(10)^{8} m[/tex] is the radius of the Sun
Simplifying:
[tex]g_{sun}=G\frac{M}{r^{2}}[/tex] (2)
[tex]g_{sun}=6.674(10)^{-11} \frac{m^{3}}{kgs^{2}} \frac{1.99(10)^{30} kg}{(6.96(10)^{8} m)^{2}}[/tex] (3)
[tex]g_{sun}=274 m/s^{2}[/tex] (4)
Since the acceleration due gravity on Earth is [tex]g_{E}=9.8 m/s^{2}[/tex], the relation is:
[tex]\frac{g_{sun}}{g_{E}}=\frac{274 m/s^{2}}{9.8 m/s^{2}}[/tex]
[tex]\frac{g_{sun}}{g_{E}}=27.9 \approx 28[/tex]
Hence, your weight would increase by a factor of 28.
The acceleration of gravity on the surface of the Sun is approximately 274 m/s^2. If you could stand on the Sun, your weight would increase by a factor of approximately 274.
Explanation:To calculate the acceleration of gravity on the surface of the Sun, we can use the formula for gravitational acceleration: g = GM/R^2. Given that the mass of the Sun (M) is 1.99 ✕ 10^30 kg, the radius of the Sun (R) is 6.96 ✕ 10^8 m, and the gravitational constant (G) is 6.67 ✕ 10^-11 N · m^2/kg^2, we can substitute these values into the formula to find the acceleration of gravity on the surface of the Sun. Using the formula, the acceleration of gravity on the surface of the Sun is approximately 274 m/s^2.
To calculate the factor by which your weight would increase if you could stand on the Sun, we can use the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2 and the weight of a 1.0-kg object is 9.8 N. Since the acceleration of gravity on the surface of the Sun is 274 m/s^2, we can calculate the weight on the Sun by multiplying the mass by the acceleration of gravity on the Sun. Dividing the weight on the Sun by the weight on Earth will give us the factor by which your weight would increase if you could stand on the Sun. Using this calculation, the weight on the Sun would be approximately 274 times greater than the weight on Earth.
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A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.
Answer:[tex]1.95 m/s^2[/tex]
Explanation:
Given
inclination [tex]\theta =30.6^{\circ}[/tex]
coefficient of kinetic friction [tex]\mu =0.36 [/tex]
As crate is moving Down therefore friction will oppose the motion
using FBD
[tex]mg\sin \theta -f_r=ma [/tex]
[tex]f_r=\mu N[/tex]
[tex]f_r=\mu mg\cos \theta [/tex]
[tex]mg\sin \theta -\mu mg\cos \theta =ma[/tex]
[tex]a=g\sin \theta -\mu g\cos \theta [/tex]
[tex]a=g(\sin (30.6)-0.36\cdot \cos (30.6))[/tex]
[tex]a=9.8\times 0.199[/tex]
[tex]a=1.95 m/s^2[/tex]
To determine the crate's acceleration down the ramp, we calculate and subtract the kinetic friction from the component of the gravitational force acting down the slope. We then divide the resulting net force by the mass of the crate.
Explanation:To find the acceleration of the crate, we need to note that there are two forces acting on the crate as it moves down the ramp: the force due to gravity and the force due to friction. Since these two forces act in opposite directions, the crate's acceleration will be less than if there were no friction.
First, we need to calculate the component of the gravitational force acting down the slope, which can be found using the formula mg sin θ, where m is the mass of the crate, g is the acceleration due to gravity, and θ is the angle of the incline.
Next, we calculate the kinetic friction using the formula μk mg cos θ, where μk is the coefficient of kinetic friction.
The net force acting on the crate is the difference between these two forces and since F = ma, where F is the net force and m is the mass of the crate, we can solve for acceleration (a) by dividing the net force by the mass of the crate.
This calculation will give us the acceleration of the crate down the incline, taking into account the presence of friction.
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Grandfather clocks are designed so they can be adjusted by moving the weight at the bottom of the pendulum up or down. Suppose you have a grandfather clock at home that runs fast. Which of the following adjustments of the weight would make it more accurate?
A. Raise the weight.
B. Decrease the amplitude of swing by a small amount.
C. Add more mass to the weight.
D. Lower the weight.
E. Remove some mass from the weight.
To fix a grandfather clock that runs fast, you should lower the weight on the pendulum (option D), as this will increase the pendulum's period and slow down the clock's ticking rate.
Explanation:If you have a grandfather clock at home that runs fast, you should make the pendulum swing slower to correct the time. To do this, you would lower the weight on the pendulum. The period of a pendulum, which determines the tick-tock rate of the clock, is directly affected by the length of the pendulum, not the weight, with a longer pendulum resulting in a slower tick-tock. Therefore, the correct adjustment would be option D: Lower the weight.
Adding or removing mass does not affect the period, so options C and E would not be effective. While decreasing the amplitude of swing might have a very small effect to speed up the pendulum (option B), this is not the standard method for adjusting the timekeeping of a clock and would not provide a consistent solution. Thus, the most accurate way to adjust the clock is by changing the length of the pendulum directly by lowering the weight.
Starting with an initial value of P(0) = 75, the population of a prairie dog community grows at a rate of
P(t) =50 − t/5 (in units of prairie dogs/month), for 0 <_t <_250.
a. What is the population 99 months later?
b. Find the population P(t) for 0 <_t <_250.
An individual experiences a deep cut that severs the radial artery near the elbow. Ignoring air resistance, approximately how high will the blood spurt? (Hints: the specific gravity of blood is 1.050 g cm-3 and the specific gravity of mercury is 13.6 g cm-3 )
The blood will spurt to a height of 0.077 meters or 7.7 centimeters.
Height of blood spurtThe height to which blood spurts from a severed artery can be calculated using Torricelli's law, which is based on the principles of fluid dynamics.
The formula for the height h that blood spouts from the artery can be derived from Torricelli's law:
h = (2 * ΔP) / (ρ * g),
where:
ΔP is the change in pressure,ρ is the density of the fluid (blood),g is the acceleration due to gravity.To find ΔP, we can calculate the pressure difference between the cut artery and the atmosphere using the following formula:
ΔP = ρ * g * h.
Given that the specific gravity of blood is 1.050 g cm⁻³, we can find its density:
ρ_blood = [tex](1.050 g cm^{-3}) * (1 g / cm^3) = 1.050 g / cm^3.[/tex]
The specific gravity of mercury is 13.6 g[tex]cm^{-3[/tex], which corresponds to its density:
ρ_mercury = 13.6 g / [tex]cm^3[/tex].
Now, we can find the pressure difference:
ΔP = [tex](1.050 g / cm^3) * (9.81 m/s^2) * h[/tex],
where g is the acceleration due to gravity (approximately 9.81 m/s²).
We'll assume that the height (h) is in meters. We want to find h, so we'll solve for it:
h = ΔP / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
Now, plug in the values:
h = ΔP / ((1.050 g / cm³) * (9.81 m/s²)).
h = (ρ_mercury * g * h) / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
Now, solve for h:
h = (13.6 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex]) * h / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
h = (13.6 / 1.050) * h.
h = 12.952 * h.
h = h / 12.952.
1 = 1 / 12.952.
h = 1 / 12.952.
h ≈ 0.077 m.
So, the blood will spurt to a height of approximately 0.077 meters or 7.7 centimeters.
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The blood spurt from a deep cut is determined by the blood pressure, which needs to exceed the vein pressure for fluid entry, and by the pressure gradient from the heart to smaller vessels. These factors, combined with the density disparity between blood and mercury, influence the height of the blood spurt.
Explanation:To determine how high the blood will spurt when an individual experiences a deep cut that severs the radial artery near the elbow, we first need to understand the pressure at which the fluid, in this case blood, enters the vascular system. It must exceed the blood pressure in the vein, which is approximately 18 mm Hg above atmospheric pressure.
The blood pressure of a young adult is represented by 120 mm Hg at systolic (maximum output of the heart) and 80 mm Hg at diastolic (due to the elasticity of arteries maintaining pressure between heartbeats). Thus, the density of mercury which raises the mercury fluid in the manometer is 13.6 times greater than water, meaning the height of the fluid will be 1/13.6 of that in a water manometer.
The blood leaves the heart with a pressure of about 120 mm Hg but its pressure continues to decrease as it goes from the aorta to smaller arteries to small veins. The pressure differences in the circulation system are due to the blood flow as well as the position of the person.
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A block of mass m attached to a horizontally mounted spring with spring constant k undergoes simple harmonic motion on a rictionless surface. How would the maximum speed of the block be affected i the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant?
1) It would remain unchanged.
2) It would increase by a factor of 2.
3) It would decrease by a factor of 1/4
4) It would increase by a factor of 4.
5) It would decrease by a factor of 1/2
To solve this problem it is necessary to apply the concepts related to simple harmonic movement.
The maximum speed from the simple harmonic motion is given as
[tex]V = A\sqrt{\frac{K}{m}}[/tex]
Where,
K = Spring constant
m = mass
At this case m is a constant then
[tex]V \propto \sqrt{K}[/tex]
then the ratio is given by
[tex]\frac{v_2}{v_1}=\sqrt{\frac{K_1}{K_2}}[/tex]
According the statement,
[tex]v_2 = \sqrt{\frac{K_1}{K_2}}v_1[/tex]
[tex]v_2 = 2v_1[/tex]
Therefore the maximum speed becomes double: 2) It would increase by a factor of 2.
Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?
The measurement will be significantly affected.
Recall that the relationship between linear velocity and angular velocity is subject to the formula
[tex]v = \omega r[/tex],
Where r indicates the radius and [tex]\omega[/tex] the angular velocity.
As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.
Mounting larger-diameter tires on a truck leads to a systematic error in the speedometer reading, whereby the speedometer will underreport the truck's true linear speed due to an increase in linear distance covered per tire rotation at a constant angular velocity.
If larger-diameter tires are mounted on a truck, the speedometer reading will not be correct. This is due to the relationship between the angular velocity of the tires and the linear speed of the truck. Since the speedometer measures the angular speed of the tires to determine the truck's linear speed, changing to larger tires will cause a systematic error in the reading. For a fixed angular velocity, larger tires will cover a greater linear distance because of their larger circumference. Thus, the speedometer will display a speed lower than the truck's true linear speed. This affects the accuracy of the speedometer.
To further illustrate, consider a truck tire rotating with an angular velocity α. The linear (tangential) velocity v at the surface of the tire is given by v = rα, where r is the tire radius. With larger tires, for the same angular velocity α, r is bigger, hence v is larger. Therefore, when the tires are larger, the speedometer, which is calibrated for the standard tire size, underestimates the truck's speed because it is based on an incorrect assumption about the tire circumference.
What is mass wasting? Select one: a. Chemical and physical weathering of dirt and rock b. Movement of regolith down a slope under the force of gravity c. Weathering of regolith by rainfall d. Oversaturation of loose rock and soil by water e. Flash flooding by a stream
Answer:
B. Movement of regolith down a slope under the force of gravity
It is also known as slope movement or mass movement.
In moving out of a dormitory at the end of the semester, a student does 1.82 x 104 J of work. In the process, his internal energy decreases by 4.07 x 104 J.
Determine each of the following quantities (including the algebraic sign):
(a)W,
(b)U
(c)Q.
Answer:
Explanation:
(a) Work done, W = 1.82 x 10^4 J
(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)
(c) According to the first law of thermodynamics
Q = W + U
Q = 1.82 x 10^4 - 4.07 x 10^4
Q = - 2.25 x 10^4 J
The work done (W), the change in internal energy (ΔU), and heat transferred (Q) are 1.82 x 10^4 J, -4.07 x 10^4 J, and -2.25 x 10^4 J respectively as per the First Law of Thermodynamics.
Explanation:The values listed in the question are for the work done (W), and the change in internal energy (U). In physics, these concepts are investigated under a principle called the First Law of Thermodynamics, which is basically a version of the law of conservation of energy as applied for thermal processes. As per this law, the change in internal energy (U) of a system is equal to the heat added to the system (Q) minus the work done by the system (W).
Hence, the formula can be given as: ΔU = Q - W.
If we substitute the given values:
-W = Work done = 1.82 x 10^4 J and ΔU = internal energy decreases = -4.07 x 10^4 J
Now, the formula can be rewritten as:
:-Q = ΔU + W = -4.07 x 10^4 J + 1.82 x 10^4 J = -2.25 x 10^4 J
Values:
(a) W = Work done = 1.82 x 10^4 J (Work done by the system is positive.)
(b) ΔU = Change in internal energy = -4.07 x 10^4 J (Decrease in internal energy is negative.)
(c) Q = Heat transferred = -2.25 x 10^4 J (As per the convention, heat added to the system is positive. Here, it's negative, which means heat is lost from the system.)
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In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 52 m/s (terminal speed), that his mass (including gear) was 86 kg, and that the force on him from the snow was at the survivable limit of 1.2 ✕ 105 N.What is the minimum depth of snow that would have stopped him safely?
Answer:
s = 0.9689 m
Explanation:
given,
Height of fall of paratroopers = 362 m
speed of impact = 52 m/s
mass of paratrooper = 86 Kg
From from snow on him = 1.2 ✕ 10⁵ N
now using formula
F = m a
a = F/m
[tex]a = \dfrac{1.2 \times 10^5}{86}[/tex]
[tex]a =1395.35\ m/s^2[/tex]
Using equation of motion
v² = u² + 2 a s
[tex]s =\dfrac{v^2}{2a}[/tex]
[tex]s =\dfrac{52^2}{2\times 1395.35}[/tex]
s = 0.9689 m
The minimum depth of snow that would have stooped him is s = 0.9689 m
Objects with masses of 130 kg and a 430 kg are separated by 0.300 m.(a) Find the net gravitational force exerted by these objects on a 35.0 kg object placed midway between them.(b) At what position (other than infinitely remote ones) can the 35.0 kg object be placed so as to experience a net force of zero?
Final answer:
The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, and for a 35.0 kg object to experience a net force of zero, it must be at a point where gravitational pulls from surrounding masses are balanced.
Explanation:
To solve the student’s question regarding the gravitational forces, we will use Newton's Law of Universal Gravitation which states the gravitational force (F) between two masses (m1 and m2) is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers. It is mathematically expressed as F = G * (m1 * m2) / r², where G is the gravitational constant (6.674 × 10-11 N·m²/kg²).
(a) To find the net gravitational force exerted on the 35.0 kg object placed midway between the two objects with masses of 130 kg and 430 kg and separated by 0.300 m, we calculate the gravitational force from each mass separately and then find the vector sum which in this case will just be the difference, as the object is placed in the middle.
(b) For the 35.0 kg object to experience a net force of zero, it has to be placed at a point where the gravitational force due to both masses equals each other. This can be found by setting the gravitational forces from each mass to the 35.0 kg object equal and solving for the distance from one of the masses. There is only one point between the two masses where this can occur as per the symmetry of the system.
A typical meteor that hits the earth’s upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at an impressive 40 km/s. As the meteor slows, the resulting thermal energy makes a glowing streak across the sky, a shooting star. The small mass packs a surprising punch. At what speed would a 900 kg compact car need to move to have the same kinetic energy
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor [tex]m=2.5 gm\approx 2.5\times 10^{-3} kg[/tex]
velocity of meteor [tex]v=40km/s \approx 40000 m/s[/tex]
Kinetic Energy of Meteor
[tex]K.E.=\frac{mv^2}{2}[/tex]
[tex]K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}[/tex]
[tex]K.E.=2\times 10^6 J[/tex]
Kinetic Energy of Car
[tex]=\frac{1}{2}\times Mu^2[/tex]
[tex]=\frac{1}{2}\times 900\times u^2[/tex]
[tex]\frac{1}{2}\times 900\times u^2=2\times 10^6 [/tex]
[tex]900\times u^2=4\times 10^6[/tex]
[tex]u^2=\frac{4}{9}\times 10^4[/tex]
[tex]u=\frac{2}{3}\times 10^2[/tex]
[tex]u=66.67 m/s[/tex]
Answer:
v = 67 m/s
Explanation:
The meteor has a mass (m) of 2.5 g and a speed (v) of 40 km/s. In SI units:
2.5 g × (1 kg / 10³ g) = 2.5 × 10⁻³ kg
40 km/s × (10³ m / 1 km) = 4.0 × 10⁴ m/s
The kinetic energy (KE) is:
KE = 1/2 × m × v² = 1/2 × (2.5 × 10⁻³ kg) × (4.0 × 10⁴ m/s)² = 2.0 × 10⁶ J
A 900 kg compact car, with the same kinetic energy, must have the following speed.
KE = 1/2 × m × v²
2.0 × 10⁶ J = 1/2 × 900 kg × v²
v = 67 m/s
A curve of radius 80 m is banked at 45 deg.
Suppose that an ice storm hits, and the curve is effectively frictionless.
What is the safe speed with which to take the curve without either sliding up or down?
Answer:
[tex]v=28m/s[/tex]
Explanation:
The vertical component of the normal force must cancel out with the weight of the car taking the curve:
[tex]N_y=W[/tex]
[tex]Ncos\theta=mg[/tex]
(Notice it has to be cos and not sin, because the angle [tex]\theta[/tex] is the slope, for null slope [tex]N_y=Ncos(0)=N[/tex], as it should be).
The horizontal component of the normal force must be the centripetal force, that is:
[tex]N_x=F_{cp}[/tex]
[tex]Nsen\theta=ma_{cp}[/tex]
[tex](\frac{mg}{cos\theta})sin\theta=m\frac{v^2}{r}[/tex]
[tex]gtan\theta=\frac{v^2}{r}[/tex]
[tex]v=\sqrt{grtan\theta}=\sqrt{(9.8m/s^2)(80m)tan(45^{\circ})}=28m/s[/tex]
To find the safe speed on a banked curve without sliding up or down, calculate the square root of a specific formula involving the radius, acceleration due to gravity, and the tangent of the angle.
Explanation:To find the safe speed with which to take the curve without sliding up or down, we need to consider the force components acting on the vehicle. The gravitational force can be split into two components: the vertical component and the horizontal component. The vertical component helps keep the vehicle on the road, while the horizontal component provides the centripetal force needed to keep the vehicle moving in a curved path. By setting up an equation involving the gravitational force, the normal force, and the friction force, we can calculate the safe speed as the square root of the product of the radius of the curve, the acceleration due to gravity, and the tangent of the banking angle.
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The energy in an oscillating LC circuit containing a 1.57 H inductor is 5.76 μJ. The maximum charge on the capacitor is 201 μC. For a mechanical system with the same period, find the
(a) mass
(b) spring constant
(c) maximum displacement
(d) maximum speed.
Answer:
(a)1.57 kg
(b) 281.17 N/m
(c) 201 micrometer
(d) [tex]2.69\times 10^{-3}m/sec[/tex]
Explanation:
We have given that value of inductor L = 1.57 Henry
Inductive energy [tex]E=5.76\mu j=5.76\times 10^{-6}J[/tex]
Maximum charge [tex]Q=201\mu C=201\times 10^{-6}C[/tex]
(A) In electrical mechanical system mass corresponds to inductance
So mass will be m = 1.57 kg
(B) We have given energy [tex]E=\frac{Q^2}{2C}[/tex]
[tex]C=\frac{Q^2}{2E}=\frac{(201\times 10^{-6})^2}{2\times 5.7\times 10^{-6}}=3543.94\times 10^{-6}[/tex]
In electrical mechanical system spring constant is equivalent to [tex]\frac{1}{C}[/tex]
So spring constant [tex]k=\frac{1}{C}=\frac{1}{3543.94\times 10^{-6}}=282.17N/m[/tex]
(c) Displacement is equivalent to maximum charge
So displacement will be [tex]x=201\mu m[/tex]
(d) Maximum speed is correspond to maximum current
As maximum current [tex]i_m=\frac{Q}{\sqrt{LC}}=\frac{201\times 10^{-6}}{\sqrt{1.57\times 3543.94\times 10^{-6}}}=2.69\times 10^{-3}A=2.69\times 10^{-3}m/sec[/tex]
A student stands on a horizontal platform that is free to rotate around a vertical axis. He holds two weights in his outstretched hands. Another student gives him a push and starts the platform rotating at 0.500 rev/s. The student then pulls the weights in close to his chest. The moment of inertia with the weights extended is 2.40 kgrn2; the moment of inertia with the weights close to the axis is 0.904 kgm2. Ignore any frictional effects. What is his new rate of rotation?
Answer:
1327.43362 rev/s
Explanation:
[tex]I_i[/tex] = Initial moment of inertia = 2.4 kgm² (arms outstretched)
[tex]I_f[/tex] = Final moment of inertia = 0.904 kgm² (arms close)
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity = 0.5 rev/s
Here the angular momentum is conserved
[tex]I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\frac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\frac{2.4\times 500}{0.904}\\\Rightarrow \omega_f=1327.43362\ rev/s[/tex]
The new rate of rotation is 1327.43362 rev/s
A railroad tank car contains milk and rolls at a constant speed along a level track. The milk begins to leak out the bottom. The car then
a. Need more information about the rate of the leak.
b. maintains a constant speed.
c. speeds up.
d. slows down.
The answer is option d. slows down.
Explanation:The answer to the question is option d. slows down.
When the milk begins to leak out of the bottom of the railroad tank car, the car will experience a decrease in its mass as the milk is being lost. According to the Law of Conservation of Momentum, if the mass of an object decreases without any external forces acting on it, the object will slow down.
Therefore, as the milk leaks out of the tank car, its speed will slow down over time.