Work = (force) x (distance)
Your force was not enough to move the building, so the distance is zero.
Work = (your force) x (zero)
Work = zero
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of 128 kg. Under the influence of a drive force of 195 N, it is moving at a constant velocity whose magnitude is 5.90 m/s. The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.
Answer:
a) Δx = 11.6 m
b) t = 3.9 s
Explanation:
a)
Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:[tex]a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)[/tex]
Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:[tex]v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)[/tex]
Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:[tex]\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)[/tex]
b)
We can find the time needed to come to an stop, applying the definition of acceleration, as follows:[tex]v_{f} = v_{o} + a*\Delta t (4)[/tex]
Since we have already said that the snowmobile comes to an stop, this means that vf = 0.Replacing a and v₀ as we did in (3), we can solve for Δt as follows:[tex]\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)[/tex]
PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE
Which best explains why species living in Australia are found nowhere else on Earth? This is an example of Geologic Evolution.
A.
Australia has an ecosystem different from any other area on Earth.
B.
Humans have genetically altered many Australian species in laboratories.
C.
Australian species were genetically altered after a comet hit the landmass.
D.
Australia separated from other continents and species there evolved independently.
In a magnetized object, most of the domains point
a. in the same direction.
b. in an east-west direction.
c. in different directions.
d. perpendicular to one another.
Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed
Answer:
The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg
Explanation:
The given percentage by weight of protein solids in raspberries = 10 weight%
The ratio of sugar to raspberries in ja-m = 45:55
The mass of the mixture after boiling = 0.4 weight fraction water
Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry
The mass of raspberry, r = 1 kg
The percentage by weight of water in raspberry = 90 weight %
The mass of water in 1 kg of raspberry = 90/100 × 1 kg = 0.9 kg
The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55
∴ s = 1 kg × 55/45 = 11/9 kg
The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg
The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction
Let 'w' represent the mass of water boiled off, we have;
(0.9 - w)/(20/9 - w) = 0.4
(0.9 - w) = 0.4 × (20/9 - w)
0.9 - w = 8/9 - 0.4·w
9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w
(81 - 80)/(90) = (6/10)·w
1/90 = (6/10)·w
w = ((10/6) × 1/90) = 1/54
w = 1/54
The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg
The theorist who expanded the behavioristic perspective to include cognitive influences on personality was
Please select the best answer from the choices provided
O Sigmund Freud
O Albert Bandura
OB. F. Skinner
O Carl Jung
Answer:
It is actually Albert Bandura! Hope this helps<3
Explanation:
In contrast to Skinner's idea that the environment alone determines behavior, Bandura (1990) proposed the concept of reciprocal determinism, in which cognitive processes, behavior, and context all interact, each factor simultaneously influencing and being influenced by the others.
The theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.
What is Personality?Personality is defined as a distinctive set of behavior, cognition, and emotional patterns that are formed by biological and environmental factors, and that change over time. This explained as the enduring characteristics and behavior which comprise a person's unique adjustment to life, including major traits, interests, drives, values, self-concept, abilities, and emotional patterns.
The four theory is also called as the proto-psychological theory which suggests that there are four fundamental personality types: sanguine, choleric, melancholic, and phlegmatic. Albert Bandura (1990) proposed the concept of reciprocal determinism in which cognitive processes, behavior, and context all interact, with each factor simultaneously affecting and being affected by the others.
Thus, the theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.
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Based on the definition of weather, ____________, is the basic difference climate and weather.
a)precipitation
b)condensation
C)time
D)jet stream
Answer:
The answer is C. time. Please mark as brainliest.
A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun
The parallel plates in a capacitor, with a plate area of 6.70 cm2 and an air-filled separation of 2.50 mm, are charged by a 8.70 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 9.90 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
Answer:
a. 34.45 V
b. 89.80 pJ
c. 355.57 pJ
d. -265.67 pJ
Explanation:
(a) the potential difference between the plates
Since charge is conserved, charge before separation, Q = charge after separation, Q'
Q = C₁V₁ and C₁ = capacitance before separation = ε₀A/d₁ where A = area of plates = 6.70 cm² = 6.70 × 10⁻⁴ m²and d₁ = initial separation of plates = 2.50 mm = 2.50 × 10⁻³ m, V₁ = initial potential difference across plates = 8.70 V
Q' = C₂V₂ and C₂ = capacitance before separation = ε₀A/d₂ where A = area of plates = 6.70 cm² = 6.70 × 10⁻⁴ m² and d₁ = final separation of plates = 9.90 mm = 9.90 × 10⁻³ m. V₂ = final potential difference across plates = unknown
So, Q = Q'
C₁V₁ = C₂V₂
ε₀AV₁/d₁ = ε₀AV₂/d₂
So, V₁/d₁ = V₂/d₂
V₂ = V₁d₂/d₁
Substituting the values of the variables into the equation, we have
V₂ = 8.70 V × 9.90 mm/2.50 mm
V₂ = 8.70 V × 3.96
V₂ = 34.452 V
V₂ ≅ 34.45 V
(b) the initial stored energy
The energy stored in a capacitor U = 1/2CV².
So, the initial energy stored U₁ = 1/2C₁V₁² = ε₀AV₁²/2d₁
Substituting the values of the variables into the equation, we have
U₁ = ε₀AV₁²/2d₁
U₁ = 8.854 × 10⁻¹² F/m × 6.70 × 10⁻⁴ m² × (8.70V)²/(2 × 2.50 × 10⁻³ m)
U₁ = 4490.07 × 10⁻¹⁶ FmV²/5 × 10⁻³ m
U₁ = 898.01 × 10⁻¹³ J
U₁ = 89.801 × 10⁻¹² J
U₁ = 89.801 pJ
U₁ ≅ 89.80 pJ
(c) the final stored energy
The energy stored in a capacitor U = 1/2CV².
So, the initial energy stored U₂ = 1/2C₂V₂² = ε₀AV₂²/2d₂
Substituting the values of the variables into the equation, we have
U₂ = ε₀AV₂²/2d₂
U₂ = 8.854 × 10⁻¹² F/m × 6.70 × 10⁻⁴ m² × (34.45 V)²/(2 × 9.90 × 10⁻³ m)
U₂ = 70403.26 × 10⁻¹⁶ FmV²/19.8 × 10⁻³ m
U₂ = 3555.72 × 10⁻¹³ J
U₂ = 355.572 × 10⁻¹² J
U₂ = 355.572 pJ
U₂ ≅ 355.57 pJ
(d) the work required to separate the plates.
The work required to separate the plates W = -ΔU is the difference between the energy stored at the separation of 9.90 mm and 2.50 mm.
So, W = -(U₂ - U₁) = -(355.57 pJ - 89.80 pJ) = -265.67 pJ.
A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding
Answer:
0.6
Explanation:
Given that :
Radius, R = 7m
Period, T = 6.9s
The Coefficient of static friction, μs can be obtained using the relation :
μs = v² / 2gR
Recall, v = 2πR/T
μs becomes ;
μs = (2πR/T)² / 2gR
μs = (4π²R² / T²) ÷ 2gR
μs = (4π²R² / T²) * 1/ 2gR
μs = 4π²R / T²g
μs = 4π²*7 / 6.9^2 * 9.8
μs = 28π² / 466.578
μs = 276.34892 / 466.578
μs = 0.5922887
μs = 0.6
How much heat is needed to boil 5.30 kg of water at its boiling point?
Use Q = mass x latent heat of vaporization.
A. 1760 kJ B. 22.2 kJ C. 530 kJ D. 12,000kJ
Answer:
Required heat Q = 11,978 KJ
Explanation:
Given:
Mass = 5.3 kg
Latent heat of vaporization of water = 2,260 KJ / KG
Find:
Required heat Q
Computation:
Required heat Q = Mass x Latent heat of vaporization of water
Required heat Q = 5.3 x 2260
Required heat Q = 11,978 KJ
Required heat Q = 12,000 KJ (Approx.)
Consider taking a time-lapse video of an analog clock that is missing the hour hand. Assume that one frame of video is taken every X minutes. When you view the compiled video, it appears as though the minute hand is advancing 10 minutes every frame. (5 points/each) What is the frame rate of the time-lapse video if no aliasing occurs
Answer:
the frame rate of the time-lapse video if no aliasing occurs is 10min per frame
Explanation:
Given the data in the question;
No aliasing effect, meaning the clock with missing hour hand will not have any effect
Time required for one frame = 10 minutes
frame rate of time = Time required / number of frame
= 10 min / 1 frame
= 10min per frame
Therefore, the frame rate of the time-lapse video if no aliasing occurs is 10min per frame
A toy projectile is fired vertically from the ground upward with a velocity of +29 meters per second. It arrives at its maximum altitude in 3.0 seconds. How high does the projectile go?
Answer:
[tex]\boxed{\text{\sf \Large 42 m}}[/tex]
Explanation:
Use height formula
[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]
u is initial velocity
θ = 90° (fired vertically upward)
g is acceleration of gravity
[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]
Suppose you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 50 mi/hr with a constant 5-second time headway between them. All vehicles in lane 2 are traveling at 55 mi/hr with a constant 6-second time headway between them. All vehicles in lane 3 are traveling at 60 mi/hr with a constant 10-second time headway between them. You collected spot speed data for all vehicles as they crossed your observation point for 75 minutes. After 10 minutes, vehicles in lane 1 stopped arriving, and after 30 minutes, vehicles in lane 2 stopped arriving. What is the flow and density of the observed traffic stream
Answer:
lane 3 Ф = 450 vehicles, ρ = 0.1 vehicle / s
lane 2 Ф_{average} = 300 vehicles, ρ _{average} = 6.66 10⁻² vehicles/s
lane 1 Ф_{average} = 300 vehicles, ρ_{average} = 2.66 10⁻² vehicle/s
Explanation:
Before solving this exercise we must clarify the concepts the flow is defined as the occurrence of an event in a time interval, in this case the passage of a car through time
Flux Density is the flux between unit area or unit time
Let's start by calculating the calculation for lane 3
the flow.
Let's use a direct rule of proportions (rule three) if the number of vehicles per unit of time (t₀ = 10s), for the observation time how many vehicles passed in the observation time (t_total = 75 * 60 = 4500 s)
Ф = 4500 s (1 vehicle / 10 s)
Ф = 450 vehicles
The flux density is the flux per unit area, in this case the area is not indicated, so we can define the flux density as the flux per unit of time.
ρ = 450/4500
ρ = 0.1 vehicle / s
Lane 2
we look for the flow
we can have separates the interval into two parts
* for the first t₁1 = 30 * 60 = 1800 s
Ф₁ = 1800 s (1 vehicle / 6s)
Ф₁ = 300 vehicles during t₁
* for the rest of the time t₂ = 4500-1800 = 2700 s
Ф₂ = 0
the average density is the total number of vehicles between the total time
#_ {vehicle} = 300 +0
Ф_{average) = # _vehicle
Ф_{average} = 300 vehicles in all time
The density is
ρ 1 = fi1 / t1
ρ1 = 300/1800
ρ1 = 1.66 10-1 vehicles / s
the average density is
ρ_{average} = [tex]\frac{\phi_1 + \phi_2}{ t_{total}}[/tex]
ρ _{average} = (300 +0) / 4500
ρ _{average} = 6.66 10⁻² vehicles / s
Lane 1
flow
* first time interval t₁ = 10 * 60 = 600 s
Ф₁ = 600 s (1 vehicle / 5s)
Ф₁ = 120 vehicles in interval t₁
* second interval t₂ = 4500-600 = 3900 s
Ф2 = 0
average flow
Ф = Ф1 + Ф2
Ф = 120 vehicles at all time
Density
* first interval
ρ₁ = 120/600
ρ₁ = 0.2 vehicles / s
* second interval
ρ₂ = 0
average density
ρ+{average} = 120/4500
ρ_{average} = 2.66 10⁻² vehicle/s
100 POINTS!!!!
Use complete sentences to explain how breathing is affected in high altitudes.
Answer:
higher altitude make it harder for people to breathe
Explanation:
If I jump 4.8m/s South onto a sitting boat, what velocity do the boat and I travel at
when I land if I am 140kg and the boat is 160kg?
Answer:
An 85.0 kg fishermen jumps from a dock into an135 kg boat at rest. If the velocity of the fisherman as he leaves the dock is 4.30 m/s west, what is the final velocity of the boat and fisherman together?
A wood fire at center has red arrows labeled A pointing upward to a pair of hands. A red arrow labeled C points up and out toward a fire poker held by hands. The poker is grey except the tip, which is orange. Red arrows labeled B point outward to a pair of hands.
In which scenario does radiation occur?
A
B
C
Answer:
Its B for whoever needs it.
Answer: B
Explanation: there ya go shawtay :)
A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?
Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
HELP PLEASE DUE IN 3 MINUTES
Answer:
Tectonic Plate Movement
Explanation:
Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.
Answer:
tectonic plates movement
What happens to voltage if the resistance goes up
Answer:
This means that if the voltage is high the current is high, and if the voltage is low the current is low. Likewise, if we increase the resistance, the current goes down for a given voltage and if we decrease the resistance the current goes up.
A system consists of a copper tank whose mass is 13 kg, 4 kg of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27oC and the temperature of the water is 50oC. The electrical resistor transfers 120 kJ of energy to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature (in oC).
Answer:
T = 30.42°C
Explanation:
According to the conservation of energy principle:
[tex]Energy\ Given\ by\ Resistor = Heat\ Gain\ by\ Copper + Heat\ Gain\ by\ Water\\E = m_{c}C_{c}(T_{2c} - T_{1c}) + m_{w}C_{w}(T_{2w} - T_{1w})[/tex]
E = 120 KJ
mc = mass of copper = 13 kg
Cc = specific heat capacity of copper = 0.385 KJ/kg.°C
T2c = T2w = Final Equilibrium Temperature = T = ?
T1c = Initial Temperature of Copper = 27°C
T1w = Initial Temperature of Water = 50°C
mw = mass of water = 4 kg
Cw = specific heat capacity of water = 4.2 KJ/kg.°C
Therefore,
[tex]120\ KJ = (13\ kg)(0.385\ KJ/kg^oC)(T-27^oC) + (4\ kg)(4.2\ KJ/kg^oC)(T-50^oC)\\120\ KJ - 135.135\ KJ - 840\ KJ = (- 5.005T - 16.8 T)\ KJ/^oC\\T = \frac{-855.135\ KJ}{-28.105\ KJ/^oC}\\[/tex]
T = 30.42°C
Suppose you are standing in front of a flat mirror which is mounted to a vertical wall. For this problem you may suppose that your height is 1.70 m and your eyes are 12 cm below the top of your head. What is the smallest mirror that will still allow you to see the full length of your body
Answer:
The right approach is "0.85 m".
Explanation:
According to the question, the diagram will is provided below.
So that as per the diagram,
The values will be:
My height,
AO = 1.70 m
My eyes at,
AB = 12 cm
i.e.,
= 0.12 m
As we can see, the point of incidence lies between the feet as well as the eyes, then
BO = 1.58 m
Now,
⇒ [tex]O'D = \frac{1.58}{2}[/tex]
[tex]=0.79 \ m[/tex]
The point of incidence of the ray will be:
⇒ [tex]CO'=1.70-\frac{0.12}{2}[/tex]
[tex]=1.70-0.06[/tex]
[tex]=1.64 \ m[/tex]
Hence,
The smallest length of the mirror will be:
= [tex]CO'-O'D[/tex]
On substituting the values, we get
= [tex]1.64-0.79[/tex]
= [tex]0.85 \ m[/tex]
A trough is 10 meters long, 1 meters wide, and 2 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 2 meters, and base, on top, of length 1 meters). The trough is full of water (density 1000kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g
Answer:
The amount of work required to empty the trough by pumping the water over the top is approximately 98,000 J
Explanation:
The length of the trough = 10 meters
The width of the through = 1 meter
The depth of the trough = 2 meters
The vertical cross section of the through = An isosceles triangle
The density of water in the through = 1000 kg/m³
Let 'x' represent the width of the water at a depth
x/y = 1/2
∴x = y/2
The volume of a layer of water, dV, is given as follows;
dV = 10 × y/2 × dy = 5·y·dy
The mass of the layer of water, m = ρ × dV
∴ m = 1000 kg/m³ × 5·y·dy m³ = 5,000·y·dy kg
The work done, W = m·g·h
Where;
h = The the depth of the trough from which water is pumped
g = The acceleration due to gravity ≈ 9.8 m/s²
[tex]\therefore \, W \approx \int\limits^2_0 {5,000 \times y \times 9.8 \, dy} = \left[24,500\cdot y^2 \right]^2_0 = 98,000[/tex]
The work done by the pump to pump all the water in the trough, over the top W ≈ 98,000 J
What is one disadvantage of sending information over long distances
wirelessly using digital signals?
O A. The signals become weaker the farther the receiver is from the
source.
B. The farther the signals travel, the more slowly they move.
C. The signals become stronger the farther the receiver is from the
source.
O D. The farther the signals travel, the easier they are to detect.
The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."
Since most of the signal which we use for communication are radio signal Radio signal are basically electromagnetic waves.As the wave moves forward it looses its amplitude.So basically radio signal becomes weaker for long distance.What are radio signals?Radio signals or radio waves are a form of electromagnetic wave. Although this may sound complicated, it is possibly sufficient to say that these waves have both electric and magnetic components. They are the same as light rays, ultra-violet and infra-red. The only difference is in the wavelength of the waves.Thus , The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."
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The magnetic field of a magnet is strongest
a. at its north pole.
b. at its south pole.
c. in the middle.
d. at both of its poles.
2) An object tied to the end of a string moves in a circle. The force exerted by the string
depends on the mass of the object, its speed, and the radius of the circle. What
combination of these variables expresses the relation?
Answer:
Centripetal force formula
Explanation:
F = mv²
----
r
A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Explanation:
Given:
mass of truck M = 1370 kg
speed of truck = 12.0 m/s
mass of car m = 593 kg
collision is elastic therefore,
Applying law of momentum conservation we have
momentum before collision = momentum after collision
1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)
Also for a collision to be elastic,
velocity of approach = velocity of separation
12 -0 = v2-v1 ....(ii)
using (i) and (ii) we have
So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision
Answer:
v = 28.98 ft / s
Explanation:
For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision
In the exercise they indicate the weight of each car
Wₐ = 1500 lb
W_b = 1125 lb
Car B's velocity from v_b = 42.0 mph westward, car A travels east
let's find the mass of the vehicles
W = mg
m = W / g
mₐ = Wₐ / g
m_b = W_b / g
mₐ = 1500/32 = 46.875 slug
m_b = 125/32 = 35,156 slug
Let's reduce to the english system
v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s
We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved
we assume the direction to the east (right) positive
initial instant. Before the crash
p₀ = mₐ v₀ₐ - m_b v_{ob}
final instant. Right after the crash
p_f = (mₐ + m_b) v
the moment is preserved
p₀ = p_f
mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v
v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]
we substitute the values
v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]
v = 0.559 v₀ₐ - 26.40 (1)
Now as the two vehicles united we can use the relationship between work and kinetic energy
the total mass is
M = mₐ + m_b
M = 46,875 + 35,156 = 82,031 slug
starting point. Jsto after the crash
K₀ = ½ M v²
final point. When they stop
K_f = 0
The work is
W = - fr x
the negative sign is because the friction forces are always opposite to the displacement
Let's write Newton's second law
Axis y
N-W = 0
N = W
the friction force has the expression
fr = μ N
we substitute
-μ W x = Kf - Ko
-μ W x = 0 - ½ (W / g) v²
v² = 2 μ g x
v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5
v = 28.98 ft / s
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.940 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2 (c) Calculate the force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N
Answer:A)u =4.295m/s , B)a = 29.746m/s² C) F=3,153N
Explanation:
Using the kinematic expression
v² = u² - 2as
where
u = initial velocity
v = final velocity
s = distance
g = acceleration due to gravity .
Given that he reaches a height of 0.940 m above the floor,
the final velocity = 0
Here, acceleration due to gravity is acting in opposite the initial direction of motion. So, a=-9.81 m/s.
v² = u² + 2as
0² - u² = 2 (- 9.81) × 0.940
- u² = 2 × - 9.81 × 0.920
- u² = -18.4428
cancelling the minus in both sides , we have that
u² = 18.4428
u = √18.4428
u =4.295m/s
(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2
Using v² = u² + 2as
where u = initial speed of basketball player before lengthening = 0 m/s,
v = final speed of basketball player after lengthening = 4.295m/s,
a = acceleration while straightening his legs
s = distance moved during lengthening = 0.310m
v² = u² + 2as
a = (v² - u²)/2s
a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)
a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)
a = (18.4428 m²/s²/(0.62 m)
a = 29.746m/s²
c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N
Force= mass x acceleration.
F = 106 kg X 29.746m/s²
F = 3,153.076 rounded to 3,153N
10. A person is about to kick a soccer ball. Consider the leg and foot to be a single rigid body and that it rotates about a fixed axis through the knee joint center. Immediately prior to impact the leg and foot are in the vertical direction and the distal end of the foot has an acceleration of 10 g in the horizontal direction. The muscle force vector makes an angle of 15 degrees with the vertical and has a moment arm of 5 cm from the knee joint center. Assume the person is 1.7 m tall and has a mass of 75 kg. Find the force in the muscle
Answer:
i don't know but i hope you get it right
Explanation
A wave with a speed of 9 m/s and a frequency of 0.5 Hz has a λ of what?
Wave speed = (wavelength) x (frequency)
Wavelength = (wave speed) / (frequency)
Wavelength = (9 m/s) / (0.5 Hz)
Wavelength = 18 m