The magnitude of the forces acting at the top are;
[tex]\mathbf{F_{Top, \ x}}[/tex] = 132.95 N
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0
The magnitude of the forces acting at the bottom are;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]\mathbf{ F_f}[/tex] = -132.95 N
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = 784.8 N
The known parameters in the question are;
The mass of the person, m₁ = 70.0 kg
The length of the ladder, l = 6.00 m
The mass of the ladder, m₂ = 10.0 kg
The distance of the base of the ladder from the house, d = 2.00 m
The point on the roof the ladder rests = A frictionless plastic rain gutter
The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder
The location of the point the person is standing = 3 meters from the bottom
g = The acceleration due to gravity ≈ 9.81 m/s²
The required parameters are;
The magnitudes of the forces on the ladder at the top and bottom
The strategy to be used;
Find the angle of inclination of the ladder, θ
At equilibrium, the sum of the moments about a point is zero
The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C
Taking moment about the point of contact of the ladder with the ground, B gives;
[tex]\sum M_B[/tex] = 0
Therefore;
[tex]\sum M_{BCW}[/tex] = [tex]\sum M_{BCCW}[/tex]
Where;
[tex]\sum M_{BCW}[/tex] = The sum of clockwise moments about B
[tex]\sum M_{BCCW}[/tex] = The sum of counterclockwise moments about B
Therefore, we have;
[tex]\sum M_{BCW}[/tex] = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81
[tex]\sum M_{BCCW}[/tex] = [tex]F_R[/tex] × √(6² - 2²)
Therefore, we get;
2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = [tex]F_R[/tex] × √(6² - 2²)
[tex]F_R[/tex] = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95
The reaction force on the wall, [tex]F_R[/tex] ≈ 132.95 N
We note that the magnitude of the reaction force at the roof, [tex]F_R[/tex] = The magnitude of the frictional force of bottom of the ladder on the floor, [tex]F_f[/tex] but opposite in direction
Therefore;
[tex]F_R[/tex] = [tex]-F_f[/tex]
[tex]F_f[/tex] = - [tex]F_R[/tex] ≈ -132.95 N
Similarly, at equilibrium, we have;
∑Fₓ = [tex]\sum F_y[/tex] = 0
The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;
[tex]\sum F_y[/tex] = -70.0 × 9.81 - 10 × 9.81 + [tex]F_{By}[/tex]
∴ The upward force acting at the bottom, [tex]F_{By}[/tex] = 784.8 N
Therefore;
The magnitudes of the forces at the ladder top and bottom are;
At the top;
[tex]\mathbf{F_{Top, \ x}}[/tex] = [tex]F_R[/tex] ≈ 132.95 N←
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0 (The surface upon which the ladder rest at the top is frictionless)
At the bottom;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]F_f[/tex] ≈ -132.95 N →
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = [tex]F_{By}[/tex] = 784.8 N ↑
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A car's bumper is designed to withstand a 6.12 km/h (1.7-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.210 m while bringing a 810 kg car to rest from an initial speed of 1.7 m/s.
Answer:
5572.8 N
Explanation:
Applying,
F = ma.............. Equation 1
Where F = Force, m = mass of the car, a = acceleration.
We can find a by applying,
v² = u²+2as............. Equation 2
Where v = final velocity, u = initial velocity, a = acceleration, = distance.
From the question,
Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m
Substitute these value into equation 2
0² = 1.7²+2×0.21×a
a = -1.7²/(2×0.21)
a = -2.89/0.42
a = -6.88 m/s²
Also given: m = 810 kg
Substitute these value into equation 1
F = 810(-6.88)
F = -5572.8 N
Hence the force on the bumber is 5572.8 N
A capacitor is connected to an ac generator that has a frequency of 3.2 kHz and produces a rms voltage of 2.0 V. The rms current in the capacitor is 28 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.7 kHz, the rms current in the capacitor is 70 mA. What rms voltage does the second generator produce
Answer:
The rms voltage of new generator is 3.4 V.
Explanation:
f = 3200 Hz
rms voltage, V = 2 V
rms current, i = 28 mA
Now
f' = 4700 Hz
rms current, i' = 70 mA
let the new rms voltage is V'.
[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]
An auto mechanic needs to determine the emf and internal resistance of an old battery. He performs two measurements: in the first, he applies a voltmeter to the battery's terminals and reads 11.9 V;11.9 V; in the second, he applies an ammeter to the terminals and reads 16.1 A.16.1 A.
What are the battery's emf E and internal resistance r?
Answer:
Hence the battery's emf E is ε = 11.9 V.
The internal resistance is r = 0.739 ohms.
Explanation:
Now we know that
Voltage V = 11.9 V.
Current I = 16.1 A.
Hence this is an ideal voltmeter there are no current flows when the Voltmeter is applied.
ε = V + I r
∵ I = 0
ε = V
ε = 11.9 V
Then the ammeter is applied.
Let's take ( r ) to be the total resistance which is equal to internal resistance.
V = I r
r = [tex]\frac{V}{I}[/tex]
[tex]= \frac{11.9}{16.1}[/tex]
r = 0.739 ohms
The battery's emf (E) and internal resistance (r) are 11.9 Volts and 0.739 Ampere respectively.
Given the following data:
Voltage = 11.9 Volts.Current = 16.1 Amperes.To determine the battery's emf (E) and internal resistance (r):
How to calculate emf (E).For an ideal voltmeter, there isn't a flow of current and as such the current is equal to 0.
Mathematically, emf (E) is given by this formula:
[tex]E = V + IR[/tex]
Substituting the given parameters into the formula, we have;
[tex]E = 11.9 + 0R\\\\E = 11.9 + 0[/tex]
E = 11.9 Volts.
For the internal resistance (r):
Note: The total resistance is equal to internal resistance.
Applying Ohm's law, we have:
[tex]R = \frac{V}{I} \\\\R = \frac{11.9}{16.1}[/tex]
R = r = 0.739 Ampere.
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A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in diameter and has 1500 turns. When turned on, the current in the solenoid is increases linearly to 20 A in 1 second. What is the induced emf in the ring?
a) 2.0 x 10-5 v
b) 3.8 x 10-5 v
c) 1.2 x 10-3 v
d) 1.9 x 10-4 v
Answer:
the answer should be b) 3.8 x 10-5 v
Hai quả cầu kim loại nhỏ giống nhau, mang điện tích q1 = 2.10-8 C; q2 = 6.10-8 C, đặt cách
nhau một đoạn r trong không khí thì chúng đẩy nhau bằng một lực là 18.10-5 N. Cho hai quả cầu
tiếp xúc nhau rồi đưa về khoảng cách cũ thì lực tương tác giữa hai quả cầu là:
Answer:
fgggggffgggcffghhhjjkuuu of to ok with queen size of your yyyygtyttttttyyyhgggghhhhfrghjjkkkExplanation:
jjjgggyuuuuiiii hhjiiihuyyuuugggyujjhhhhggghhhjjhhhhjhhui
The lines in the emission spectrum of hydrogen result from __________.
a. energy given off in the form of visible light when an electron moves from a higher energy state to a lower energy state
b. protons given off when hydrogen burns
c. electrons given off by hydrogen as it cools
d. electrons given off by hydrogen when it burns
e. decomposing hydrogen atoms.
Answer:
Option (a) is correct.
Explanation:
The lines in the emission spectrum of hydrogen is due to the transfer of electrons form higher energy levels to the lower energy levels.
When the electrons transfer from one level of energy that is higher level of energy to the other means to the lower level of energy then they emit some photons which having the frequency or the wavelength in the visible region.
Both of these questions are the same but their answers in the answer key are different. Why?
Pascal's principle says: a A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. b The buoyant force equals the weight of the displaced fluid. c Matter must be conserved in a flowing, ideal fluid. d Energy is conserved in a flowing, ideal fluid. e A small input force causes a large output force.
Answer:
A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid.
Explanation:
Pascal's principle states that ''pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.''(Science direct).
The implication of this law is; that a change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. Hence the correct answer chosen above.
The Pascal's principle is applied in hydraulic jacks and automobile brakes.
(a) Calculate the force needed to bring a 800 kg car to rest from a speed of 85.0 km/h in a distance of 115 m (a fairly typical distance for a nonpanic stop).
N
(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), i.e. find the ratio of the force in part(b) to the force in part(a).
(force in part (b) / force in part (a))
Answer:
2Al + 2H2O + 2NaOH ⟶ 3H2 + 2NaAlO2
Chất rắn màu xám bạc của nhôm (Al) tan dần trong dung dịch, sủi bọt khí là hidro (H2).
Explanation:
Explain why the motor turns. Consider the Lorentz (magnetic) force we have discussed in class, and how this would apply here. Why did you have to remove only half the insulation on the ends of the wire
Solution :
Owing to the continuous attraction and repulsion force caused by the magnet or the electromagnet around the core of the motor produces a unidirectional torque whose direction is given by the Lorentz force, [tex]$F=q(\vec v \times \vec B)$[/tex], and thus the torque causes the rotation of the electric motor.
Removing half the insulation from the coil makes it to rotate just half a turn. When the half insulation is removed, the coil turns half and the rest of the time the connection terminates. The rest half turn will be provided by the angular momentum. Now after this half turn by the angular momentum, the connections will again be connected and again the torque will work on it to rotate the half turn. This continues and the motor rotates.
Physics question plz help ASAP
Answer:
Option D.
Explanation:
From the question given above, the following data were obtained:
Force applied (F) = 5 N
Extention (e) = 0.075 m
Spring constant (K) =?
The spring constant for the spring can be obtained as follow:
F = Ke
5 = K × 0.075
Divide both side by 0.075
K = 5 / 0.075
K = 67 N/m
Thus, the spring constant for the spring is 67 N/m
A 30-cm-diameter, 90-cm-high vertical cylindrical container is partially filled with 60-cm-high water. Now the cylinder is rotated at a constant angular speed of 180 rpm. Determine how much the liquid level at the center of the cylinder will drop as a result of this rotational motion.
Answer:
[tex]\triangle h_c =0.204m[/tex]
Explanation:
Diameter [tex]d=30cm[/tex]
Height [tex]h=90cm[/tex]
Fill height [tex]h_f=60cm[/tex]
Angular speed [tex]N=180rpm[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]\omega=\frac{2 \pi*N}{60}[/tex]
[tex]\omega=\frac{2 \pi*180}{60}[/tex]
[tex]\omega=18.85rads/s[/tex]
Generally the equation for Liquid surface is mathematically given by
[tex]\mu_s=h*\frac{\omega^2*0.15^2}{4*9.81}[/tex]
[tex]\mu_s=0.396m[/tex]
Therefore the liquid drop at center due to rotation is
[tex]\triangle h_c =h-\mu_s[/tex]
[tex]\triangle h_c =0.60-0.396[/tex]
[tex]\triangle h_c =0.204m[/tex]
A silicon solar cell behaves like a battery with a 0.46 V terminal voltage. Suppose that 1.0 W of light of wavelength 620 nm falls on a solar cell and that 50%% of the photons give their energy to charge carriers, creating a current. What is the solar cell's efficiency that is, what percentage of the energy incident on the cell is converted to electric energy?
We have that the percentage of the energy incident on the cell that is converted to electric energy is
[tex]n=11\%[/tex]
From the question we are told that:
Voltage [tex]V=0.46V[/tex]
Power of light [tex]P=1.0W[/tex]
Wavelength [tex]w=620nm[/tex]
50 \% of the photons give their energy to charge carriers,
Generally, the equation for number of Protons is mathematically given by
[tex]N_p=\frac{P}{E}[/tex]
[tex]N_p=\frac{P \lambda}{hc}[/tex]
[tex]N_p=\frac{1}{(6.62*10^(-34)}*\frac{3*10^8}{(570*10^{-9}))}[/tex]
[tex]N_p=2.87*10^{18}[/tex]
Generally, the equation for Number of electron is mathematically given by
[tex]N_e=50 \% *n_3[/tex]
[tex]N_e=0.5*2.87*10^{18}[/tex]
[tex]N_e=1.43*10^{18}[/tex]
Therefore
Total current
[tex]I= e*N_e[/tex]
Where
e=electron Charge
Therefore
[tex]I=1.43*10^{18}*1.6*10^-{19}[/tex]
[tex]I=0.230A[/tex]
Generally, the equation for Power is mathematically given by
[tex]P=VI[/tex]
[tex]P=0.46*0.230[/tex]
[tex]P=0.1058W[/tex]
Therefore
Efficiency
[tex]n=\frac{0.1058}{1}[/tex]
[tex]n=0.1058[/tex]
[tex]n=11\%[/tex]
In conclusion
The percentage of the energy incident on the cell that is converted to electric energy is
[tex]n=11\%[/tex]
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What is the incorrect statement regarding the isotopes of the same element?
1) Electronic configuration is equal
2) Mass number is equal
3) Number of protons are equal
4) Number of electrons are equal
Answer:
1231
Explanation:
A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.
Answer:
[tex]0.2677\ \text{V/m}[/tex]
Explanation:
A = Area of loop = [tex]0.129\times0.402[/tex]
B = Magnetic field = [tex]0.888\ \text{T}[/tex]
t = Time taken = [tex]0.172\ \text{s}[/tex]
Electric field is given by
[tex]E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}[/tex]
The emf induced is [tex]0.2677\ \text{V/m}[/tex].
What is the largest known star?
Answer:
UY Scuti is slightly larger than VY Canis Majoris
Explanation:
These stars are millions of miles away and cannot be seen by the naked eye.
Beetlejuice is another large star that can be seen by the eye.
What are the examples of pulley? Plz tell the answer as fast as possible plz.
Answer:
elevators
Theatre system
construction pulley
lifts
Answer:
elevator,cargo lift system
The velocity of a body is given by the equation v=a+bx, where 'x' is displacement. The unit of b is
Answer:
2bsnsnsnns181991oiwiw
A capacitor consists of two metal surfaces separated by an electrical insulator with no electrically conductive path through it. Why does a current flow in a resistor-capacitor circuit when the switch is closed?
Answer:
Displacement current flows in the dielectric material(insulated region)
Explanation:
Firstly a capacitor stores charge when a capacitor is charging (or discharging), current flows in the circuit. Also, there is no charge transfer in the dielectric material in the capacitor which is contradictory to the flow of current. Hence, displacement current is the current in the insulated region due to the changing electric flux.
Use the pressure meter to read the pressure in Fluid A at the bottom of the tank. Do not move the pressure meter. Switch to Fluid B and read the pressure in fluid B. Based on the two readings, compare the density of fluid B to the density of fluid A. Which statement is correct?
Answer:
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
Explanation:
The pressure at a depth of a fluid is
P = ρ g y
where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.
In this case the pressure for fluid A is
Pa = ρₐ g y
the pressure for fluid B is
P_b = ρ_b g y
depth y not changes as the gauge is stationary
if we look for the relationship between these pressures
[tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
therefore we see that the pressure measured for fluid B is different from the pressure of fluid A
if ρₐ < ρ_b B the pressure P_b is greater than the initial reading
ρₐ> ρ_b the pressure in B decreases with respect to the reading in liquid A
A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast
Answer:
x = 9.32 cm
Explanation:
For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation
Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar
- W l / 2 - W_{child} x + N₂ l = 0
x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex] 1)
now let's use the expression for translational equilibrium
N₁ - W - W_(child) + N₂ = 0
indicate that N₂ = 4 N₁
we substitute
N₁ - W - W_child + 4 N₁ = 0
5 N₁ -W - W_{child} = 0
N₁ = ( W + W_{child}) / 5
we calculate
N₁ = (450 + 250) / 5
N₁ = 140 N
we calculate with equation 1
x = -250 1.50 + 4 140 3) / 140
x = 9.32 cm
In a photoelectric effect experiment, it is observed that violet light does not eject electrons from a particular metal. Next, red light with the same intensity is incident on the same metal. Which result is possible
Answer:
No ejection of photo electron takes place.
Explanation:
When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.
The minimum energy required to just eject an electron is called work function.
The photo electric equation is
E = W + KE
where, E is the incident energy, W is the work function and KE is the kinetic energy.
W = h f
where. h is the Plank's constant and f is the threshold frequency.
Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
A rock is suspended by a light string. When the rock is in air, the tension in the string is 37.8 N. When the rock is totally immersed in water, the tension is 32.0 N. When the rock is totally immersed in an unknown liquid, the tension is 20.2 N. What is the Density of the unknown liquid?
When the rock is suspended in the air, the net force on it is
∑ F₁ = T₁ - m₁g = 0
where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So
T₁ = m₁g = 37.8 N
When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then
∑ F₂ = T₂ + B₂ - m₁g = 0
where B₂ is the magnitude of the buoyant force. Then
B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N
B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then
5.8 N = m₂g ==> m₂ ≈ 0.592 kg
If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was
1.00 × 10³ kg/m³ = m₂/V ==> V ≈ 0.000592 m³ = 592 cm³
and this is also the volume of the rock.
When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is
∑ F₃ = T₃ + B₃ - m₁g = 0
which means the buoyant force is
B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N
The mass m₃ of the liquid displaced is then
17.6 N = m₃g ==> m₃ ≈ 1.80 kg
Then the density ρ of the unknown liquid is
ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³
When using the lens equation, a negative value as the solution for di indicates that the image is
Answer:
The Anatomy of a Lens
Refraction by Lenses
Image Formation Revisited
Converging Lenses - Ray Diagrams
Converging Lenses - Object-Image Relations
Diverging Lenses - Ray Diagrams
Diverging Lenses - Object-Image Relations
The Mathematics of Lenses
Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens. The use of these diagrams was demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f)
I NEEED HELP IN PHYSICS PLEASE!
Answer:
in which topic you need help
A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.
Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.
The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.
Momentum is conserved, so the total momentum of the system is the same before and after the collision:
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
==>
(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'
==>
-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'
where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.
Kinetic energy is also conserved, so that
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²
==>
(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
==>
1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is
v₁' ≈ -3.11 m/s
v₂' ≈ -0.167 m/s
and take the absolute values to get the magnitudes.
If you want to instead use the masses from the "Required" section, you would end up with
v₁' ≈ -3.18 m/s
v₂' ≈ -0.236 m/s
A car is traveling at 104 km/h when the driver sees an accident 50 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
a = - 8.34 m/sec² ( deceleration or negative)
Equations for UAM ( uniformly accelerated motion) are:
vf = v₀ ± a*t and s = s₀ + v₀*t + (1/2)*a*t²
In our case, the motion is with deceleration, then
vf = v₀ - a*t and s = s₀ + v₀*t - (1/2)*a*t²
working on these equatios we get:
vf = v₀ - a*t (1) s - s₀ = v₀*t - (1/2)*a*t² (2)
v₀ - vf = a*t
t = (v₀ - vf)/a
By substitution of (1) in equation (2)
s - s₀ = v₀ * (v₀ - vf)/a - (1/2) * a* [(v₀ - vf)/a]²
s - s₀ = (v₀² - v₀*vf)/a - (1/2) * a* (1/a²)* (v₀ - vf)²
s - s₀ = 1/a * ( v₀² - v₀*vf ) - 1/a* (1/2) * (v₀ - vf)²
s - s₀ = 1/a* [ ( v₀² - v₀*vf ) - (1/2) * (v₀ - vf)²]
a * (s - s₀ ) = v₀² - v₀*vf - v₀²/2 - vf²/2 + v₀*vf
a * (s - s₀ ) = (1/2) * v₀² - (1/2)*vf²
a * (s - s₀ ) = (1/2) * ( v₀² - vf²)
We find an expression to calculate the minimum deceleration to stop the car in time to avoid crashing
s₀ = 50 meters s = 0 v₀ = 104 Km/h vf = 0
1 Km = 1000 m and 1 h = 3600 sec
v₀ = 104 Km/h = 28.88 m/sec
a = (1/2) [ (28.88)² - 0 ] / 0 - 50
a = - 8.34 m/sec² ( deceleration or negative)
An ink-jet printer steers charged ink drops vertically. Each drop of ink has a mass of 10-11 kg, and a charge due to 500,000 extra electrons. It goes through two electrodes that gives a vertical acceleration of 104 m/s2. The deflecting electric field is _____ MV/m.
Answer:
E = 1.25 MV / m
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F = q E
we substitute
q E = m a
E = m a / q
indicate there are 500,000 excess electrons
q = 500000 e
q = 500000 1.6 10⁻¹⁹
q = 8 10⁻¹⁴ C
the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²
let's calculate
E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴
E = 0.125 10⁷ V / m = 1.25 10⁶ V / m
E = 1.25 MV / m
A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. The balloon is floating at a constant height of 9.14 m above the ground.
Required:
What is the density of the hot air in the balloon?
9514 1404 393
Answer:
1.114 kg/m³
Explanation:
The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...
2910d +308 = 2910×1.22
Solving for d, we find ...
2910d = 2919(1.22) -308
d = 1.22 -308/2910
d ≈ 1.114 . . . kg/m³
The density of the hot air is about 1.114 kg/m³.
