Pregunta completa :
Un zapato de golf tiene 10 tacos cada uno con un área de 0.01 pulgadas en contacto con el piso suponga que al caminar hay un instante en que los tacos soportan el peso completo de una persona de 180 libras ¿cuál es la presión ejercida por los tacos sobre el suelo?
Responder:
12230825.435 pascal
Explicación:
Dado que:
Peso (W) = 180 libras
Número de tacos = 10
Área por montante = 0.01 in²
Área total (10 * 0.01) = 0.1 pulg²
Usando la calculadora de conversión:
180 libras = 81.647 kg
0.1 pulg² = 6.452 × 10 ^ -5 m²
Recordar :
Presión = Fuerza (Newton) / Área (m²)
Fuerza = masa * aceleración debida a la gravedad
Aceleración por gravedad = 9,8 m / s²
Presión = (81.647 * 9.8) / (6.452 * 10 ^ -5)
Presión = 800.1406 / 6.452 * 10 ^ -5
Presión = 12230825.435 pascal
Keeping in mind Newton’s First Law of Motion, if a rock is thrown it should stay in motion at a constant velocity. Why doesn’t it do that?
Momentum acts as a force to change its behavior.
You didn’t throw it hard enough.
Gravity acts as a force to change its behavior.
It needs an external force and an internal force to stay in motion.
According to Newton's First Law of Motion, the answer is that it needs an external force and an internal force to stay in motion. While it is true that momentum can act as a force to change its behavior, this does not necessarily explain why the rock does not stay in motion at a constant velocity.
It involve understanding the various forces at play when an object is in motion. In this case, the rock is subject to both internal and external forces that affect its motion. Internal forces, such as friction between the molecules within the rock, can slow it down and eventually bring it to a stop. External forces, such as air resistance and gravity, can also impact its motion.
For example, if you throw the rock with enough force, it may initially maintain its velocity due to its momentum. However, as it travels through the air, air resistance will gradually slow it down until it eventually hits the ground. Similarly, the force of gravity will also affect the rock's motion, pulling it downward towards the earth's surface. As a result, the rock will not stay in motion at a constant velocity because it is subject to a variety of forces that can impact its movement.
In summary, the reason why a rock thrown does not stay in motion at a constant velocity is due to the presence of internal and external forces that act on it. While momentum can play a role in its behavior, it is not the sole factor that determines its motion
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Derive the equation for the buoyant force from these two ideas.
The expression for the buoyant force is F = ρghs²
What is buoyant force?Buoyant force is the force exerted on a object by a liquid.
How to derive the expression for the buoyant force?Given that the pressure from the fluid at a certain depth is given by P = P₀ + ρgh where
P₀ = atmospheric pressure, ρ = density of fluid, g = acceleration due to gravity and h = depth of object in fluidWe know that the buoyant for is due to the pressure difference frm the top and bottom of the fluid.
Since
P = P₀ + ρgh
P - P₀ = ρgh
ΔP = ρgh
Now, we know that pressure difference, ΔP = F/A where
F = buoyant force and A = areaSo, making F subject of the formula, we have that
F = ΔPA
Now since the object is a perfect cube of length, s, its area at its bottom surface perpendicular to F is A = s²
So, F = ΔPA
F = ΔPs²
So, substituting ΔP into the equation, we have that
F = ΔPs²
F = ρghs²
So, the buoyant force is F = ρghs²
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The line drawn connecting the crests of several waves and perpendicular to the direction of wave motion is the _____.
The line drawn connecting the crests of several waves and perpendicular to the direction of wave motion is the: wavefront
The wavefront is an imaginary surface that is composed of all the points of a wave at any instant, this surface is defined as the line drawn connecting the crests of several waves and perpendicular to the direction of wave motion.
If the propagation velocity differs at many points of a wavefront, the direction and shape of the wavefronts may differ by the refraction.
What is a wave?In physics, it is the propagation of a perturbation of one of its properties in space, which involves a transmission of energy without displacement of mass.
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A car with two passengers traveling at 15 m/s collides with a tree. One of the passengers who is not wearing a seat belt strikes the windshield head first and comes to rest in 0.03 s. The area of contact between the head and the windshield is approximately 5x10 m² and the mass of the head is 5.4 kg. The other passenger who is wearing his seat belt comes to rest in 0.50 s. The mass of this passenger is 75 kg. The area of the seat belt in contact with this passenger is about 0.12 m². Find the average force and the force per unit area exerted on the two passengers.
In the absence of friction, a force applied to a body for a given amount of time, changes the momentum of the body
The average force per unit area exerted on the passenger with no seatbelt 5,400,000 N/m²
The average force per unit area for the passenger with seat belt 18,750 N/m²
The known parameters are;
The velocity of the car, v₀ = 15 m/s
The time it takes the passenger not wearing seat belt to strike the windshield, \(t_{ns}\) = 0.03 s
Area of contact between the head and the windshield, \(A_{ns}\) = 5 × 10⁻⁴ m²
The mass of the head, \(m_{h}\) = 5.4 kg
Time taken for the passenger wearing seat belt to come to rest, \(t_s\) = 0.50 s
The mass of the passenger wearing seatbelt, \(m_s\) = 75 kg
Area of the seat belt in contact with the passenger, \(A_s\) = 0.12 m²
Required:
To find the average force per unit area exerted on the two passengers
Solution:
According to Newton's second law on motion, force is equal to the product of mass and acceleration
\(Acceleration, \overline a = \dfrac{Change \ in \ velocity}{Elapsed \ time} = \dfrac{v - v_0}{t}\)
Therefore, the acceleration, force and average force per unit area of the passenger with no seat belt;
\(Acceleration, \overline a_{ns} = \dfrac{0 - 15 \, m/s}{0.03 \, s} = 500 \, m/s^2\)
\(F_{ns}\) = \(m_{h}\) × \(\overline a_{ns}\)
Which gives;
\(F_{ns}\) = 5.4 kg × 500 m/s² = 2,700 N
Force per unit area = Force/Area
The average force per unit area, for the passenger with no seat belt is given as follows:
\(P_{ns} = \dfrac{2,700 \, N}{5 \times 10 ^{-4} \ m^2} = 5,400,000 \ N/m^2\)
The average force per unit area exerted on the passenger with no seatbelt, \(P_{ns}\) = 5,400,000 N/m²
The acceleration, force and average force per unit area of the passenger with seat belt:\(Acceleration, \overline a_{s} = \dfrac{0 - 15 \, m/s}{0.5 \, s} = 30 \, m/s^2\)
\(F_{s}\) = \(m_{b}\) × \(\overline a_{s}\)
Which gives;
\(F_{s}\) = 75 kg × 30 m/s² = 2,250 N
Force per unit area = Force/Area
The average force per unit area, for the passenger with seat belt is therefore;
\(P_{s} = \dfrac{2,250\, N}{0.12 \ m^2} = 18, 750\ N/m^2\)
The average force per unit area for the passenger with seat belt, \(P_s\) = 18,750 N/m²
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Physical data is often used in the court system. In fact, police officers use radar to determine your speed when you are driving and admit that record in court if you are ever given a ticket. In one such instance, the police officer submitted data that goes against the claim of the driver, who is contesting his ticket in court. The police officer (who was stationary when he made the measure) claims that the driver was speeding at 44.7 m/s, but the driver claims he was only going 26.8 m/s (according to his speedometer). The speed limit was 60 mph. The driver argues that the radar gun must have been set incorrectly to record relative velocity. Is it possible that they are both telling the truth?
The driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.
The given parameters:
Speed of the driver observed by the stationary police officer, Vo = 44.7 m/sSpeed of the driver, V = 26.8 m/s.Speed limit = 60 mphThe speed limit of the driver in meter per second is calculated as follows;
\(= 60 \ \times \frac{miles}{hour} \times \frac{1609.34 \ m}{1 \ mile} \times \frac{1 \ hr}{3600 \ s}\\\\= 26.82 \ m/s\)
From the speed limit, it is obvious that the driver's speed is within the limit. Thus, we can conclude that the driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.
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The southern hemisphere experiences the coldest seasonal weather during what portion of the earth’s orbit?.
The southern hemisphere experiences cold weather when the south pole tilts away from the sun.
When is it cold in the southern hemisphere ?The months of January or February in the Northern Hemisphere and June, July, or August in the Southern Hemisphere often have the lowest seasonal average temperatures.
Aside from the Antarctic continent (which is undoubtedly quite cold! ), there are no other nations on or below the Antarctic Circle. This often means that winters in the Southern Hemisphere are warmer.
Seasons in the Southern Hemisphere are, as we all know, the polar opposite of those in the Northern. Instead, the Earth's orbital tilt of 23.5 degrees as it revolves around the Sun provides the solution.
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What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)
A.) 3.80 x 10^16
B.) 2.37 Hz
C.) 2.37 m
D.) 3.80 x 10^16 Hz
how is horsepower relative to the power use of a phone or light bulb?
Answer:
More horsepower = more power use.
Explanation:
Horsepower is just a measure of power, similar to Watts (1 horsepower = 745.7 W). So, as a simple example, a lightbulb that uses 1 hp will have a higher power use than a lightbulb that uses 2 hp.
A 50 g sample of an unknown metal is heated to 90. 0C. It is placed in a perfectly insulated container along with 100 g of water at an initial temperature of 20C. After a short time, the temperature of both the metal and water become equal at 25C. The specific heat of water is 4. 18 J/gC in this temperature range. What is the specific heat capacity of the metal? Record your answer with two significant figures. J/gC.
The specific heat capacity of the metal is approximately -0.64 J/g°C.
What is the specific heat capacity of the unknown metal given that a 50 g sample is heated to 90.0°C and placed in a container with 100 g of water at 20°C, resulting in both reaching a final temperature of 25°C?To determine the specific heat capacity of the metal, we can use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the metal.
The formula to calculate heat transfer is:
Heat (Q) = mass (m) ˣ specific heat capacity (c) ˣ temperature change (ΔT)
For the water:
Q_water = (100 g) ˣ (4.18 J/g°C) ˣ (25°C - 20°C)Q_metal = (50 g) ˣ (c_metal) ˣ (25°C - 90°C)Since the heat gained by the water is equal to the heat lost by the metal:
Q_water = Q_metalSubstituting the values, we have:
(100 g) ˣ (4.18 J/g°C) ˣ (25°C - 20°C) = (50 g) ˣ (c_metal) * (25°C - 90°C)Simplifying the equation:
(100 g) ˣ (4.18 J/g°C) ˣ (5°C) = (50 g) ˣ (c_metal) ˣ (-65°C)Solving for c_metal:
c_metal = [(100 g) ˣ (4.18 J/g°C) ˣ (5°C)] / [(50 g) ˣ (-65°C)]c_metal ≈ -0.64 J/g°CLearn more about heat capacity
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Squids propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 9 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. If the squid expels 2 kg of water out of its body with a speed of 8 m/s, what would be its own escape speed
Answer:
v_squid = - 2,286 m / s
Explanation:
This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.
Initial moment. Before expelling the water
p₀ = 0
the squid is at rest
Final moment. After expelling the water
\(p_{f}\) = M V_squid + m v_water
p₀ = p_{f}
0 = M V_squid + m v_water
c_squid = -m v_water / M
The mass of the squid without water is
M = 9 -2 = 7 kg
let's calculate
v_squid = 2 8/7
v_squid = - 2,286 m / s
The negative sign indicates that the squid is moving in the opposite direction of the water
States that if no external forces act on a group of objects, their total momentum does not change
Answer: this is true
Explanation:
if the object is moving and there is nothing stooping it, it will keep moving
A certain simple pendulum has a period on the earth of 1.70s . What is its period on the surface of Mars, where g=3.71m/s^2?
The period of the pendulum on the surface of Mars, where g = 3.71m/s², would be the same as the period on Earth, 1.70s. The pendulum on the surface of Mars has a period of roughly 3.31 seconds, according to our calculation of this expression.
A basic pendulum's period is determined by the equation T = 2(L/g), where T is the period, L is the pendulum's length, and g is the acceleration brought on by gravity.
L = (T/(2))2g can be solved for on Earth with a period of 1.70s by rearranging the equation.
Using the same equation and the Mars surface gravity, g = 3.71m/s2, we can now get the period on Mars' surface: T\(_{mars}\) = 2(L/g\(_{mars}\)).
The equation T\(_{mars}\) = 2(((T/(2))2g)/(g\(_{mars}\))) is obtained by substituting the expression for L and the surface gravity of Mars.
T\(_{mars}\) = T(g/g_Mars), to simplify the formula.
T\(_{mars}\) = 1.70s(9.81m/s2 / 3.71m/s) is the result of plugging in the data. The pendulum on the surface of Mars has a period of roughly 3.31 seconds, according to our calculation of this expression.
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Let n be the number of digits of the whole number. What is the number of operations that are executed in the code in terms of n?
The number of operations that are executed in the code in terms of n will depend on the code itself .
Code is a system of symbols and rules used to represent instructions that can be executed by a machine, such as a computer. It is used to communicate instructions to a computer or other electronic device in a manner that it can understand and act upon. Code can be written in a variety of languages and is used to create software applications, websites, and other digital content.
Generally speaking, the number of operations will be proportional to n, meaning that as the number of digits increases, the number of operations will increase as well. For example, if the code involves looping through each digit of the number, the number of operations will be n + 1, where n is the number of digits.
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One piece of evidence that can help astronomers sort out how the planets in our solar system formed is
Answer:
\(\huge\boxed{\sf Circumstellar\ discs }\)
Explanation:
Astronomers discovered matter made up of gas, dust or asteroids etc. around nearby stars. This matter is called circumstellar disc (due to it being ring-shaped). These circumstellar discs are materials out of which planets form.
Circumstellar discs are one piece of evidence that can help astronomers sort out how the planets in our solar system were formed.\(\rule[225]{225}{2}\)
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In an experiment, students were given an unknown mineral. The unknown mineral was placed in 150 ml of water. Once in the water, the volume went up to 165 ml. Then when the same object was dried, then massed on a balance. It had a mass of 225 grams. What is the density of the unknown mineral? Note: You only have to fill in the numerical answer.
Answer:
15
Explanation:
225/(165-150)=15g/cm3
What element atomic number is in Group 17 and Period 6?
\( \\ \)
The halogen elements are the six elements in Group 17 of the periodic table. Group 17 is the second column from the right in the periodic table and contains six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (As), and tennessine (Ts).
\( \\ \)
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A man weighing 800 Newtons is standing in an elevator. If the elevator rises with an acceleration of 9.8 meters per second2, the force exerted by the elevator on the man will beA) 400 NB) 800 NC) 1600 ND) 2000 N
The force is exerted by the elevator is (c) 1600N.
We are given that,
Weight = m = 800 Kg
Acceleration = a = 9.8 m/s²
Acceleration due to gravity = g = 9.8 m/s²
Force exerted by the elevator on the man = f = ?
f = mg + ma
Substituting the values we get,
f = m X (g + a)
f = 800 X (9.8 + 9.8)
f = 800 X 19.6
f = 15680 N
f ≈ 1600 N
If we round off the accelerations in this case to 10 m/s^2 then the force applied to the man by the lift is 1600 N.
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a wind turbine is an example of what kind of device
a. a generator
b. a transformer
c. a motor
d. an electromagnet
Answer:
it’s an example of a generator.
Explanation:
Answer:
Generator
Explanation:
valid
why do boys like boys
Answer:
maybe they're gay maybe they act gay so they can be around there girl crush cause a lot of girls like to hang out around gay guys
Explanation:
Answer:
nature is like that sometimes
and parts of nature are beautiful things <3
Explanation:
have a miraculous day!! <3
Some people say it is rare to get HIV from a blood transfusion in the United States. True or false? Why?
Answer:
False
Explanation:
Because there is a 90% chance of contracting HIV from a blood transfusion
The critical angle for an air-glass interface is 10.9 degrees. A ray in air is incident on the interface. The reflected ray is 100 percent polarized. The angle of refraction is closest to O 10.1 8.31 O 10.7 ○ 8.91。 ○ 9.51。
The correct answer is ○ 8.91.
The critical angle for an air-glass interface is the angle of incidence at which the refracted ray is at 90 degrees to the surface, meaning the angle of refraction is 90 degrees.
However, if the angle of incidence is greater than the critical angle, then total internal reflection occurs and no refracted ray is formed.
In this scenario, the reflected ray is 100 percent polarized, which means that it vibrates in only one plane.
The angle of refraction can be determined using Snell's Law, which states that n1sinθ1 = n2sinθ2,
where n1 and n2 are the refractive indices of the media on either side of the interface and θ1 and θ2 are the angles of incidence and refraction, respectively.
Since the incident ray is in air and the interface is with glass, we can assume that n1 = 1 and n2 = 1.5. Using Snell's Law, we can find the angle of refraction to be closest to 8.91 degrees.
Therefore, the answer is ○ 8.91.
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How are wavelength, pitch, frequency, and energy all related?
Answer:
he word that musicians use for frequency is pitch. The shorter the wavelength, the higher the frequency, and the higher the pitch, of the sound. In other words, short waves sound high; long waves sound low. ... In other words, it sounds higher
Explanation:
Show your work on the following problems.
1.
An airplane accelerates down a run-way at 3.20 m/s? for 32.8 s until is finally lifts off the ground
Determine the distance traveled before take-off.
2.A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. De
acceleration of the car and the distance traveled.
3.
A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the
acceleration of the car and the distance traveled.
4.
A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of
the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a
uniform acceleration).
The distance travelled by the airplane before takeoff is 1721 m.
2. The acceleration of the race car is 11.2 m/s², and Distance travelled by the car is 79.9 m.
3. The acceleration of the bullet is 1.62 * 10⁵ m/s².
What is the distance travelled by the airplane before takeoff?The distance travelled by the airplane before takeoff is determined using the formula:
s = ut + ¹/₂at²
u = 0; a = 3.20 m/s², t = 32.8 s
s = 0 * 32.0 + ¹/₂ * 3.2 * 32.8²
s = 1721 m
2. The acceleration of the race car is calculated as follows:
a = (v - u)/t
v = 46.1 m/s, u = 18.5 m/s, t = 2.47 s
a = (46.1 - 18.5)/2.47
a = 11.2 m/s²
Distance travelled, s = 18.5 * 2.47 + ¹/₂ * 11.2 * 2.47²
s = 79.9 m
3. The acceleration of the bullet is calculated as follows:
a = (v² - u²)/2s
where v = 0 m/s, u = 521 m/s, s = 0.840 m
a = (0² - 521²)/2 * 0.840
a = 161572.0238
a = 1.62 * 10⁵ m/s²
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what is the correct tight asymptotic bound for T(n) in 1,2,3 where T(1)=c ?
T(n) = 10T(n/10)+100n
T(n) =T(n/10)+100n
T(n) = T(n/10)+100
The correct tight asymptotic bound for T(n) in 1, 2, 3 where T(1)=c is Θ(n log n).
To find the tight asymptotic bound of T(n), we will use the Master Theorem. So, let's take a look at each recurrence relation:
1. T(n) = 10T(n/10)+100nApplying the Master Theorem: a = 10, b = 10, f(n) = 100nlogb a = log10 10 = 1 Since f(n) = Θ(n) = Θ(n1), Case 2 of the Master Theorem applies. The solution, therefore, is Θ(n log n).
2. T(n) = T(n/10)+100n Here, a = 1, b = 10, f(n) = 100nlogb a = log10 1 = 0 Since f(n) = Θ(n0) = Θ(1), Case 1 of the Master Theorem applies. The solution, therefore, is Θ(n).
3. T(n) = T(n/10)+100 Here, a = 1, b = 10, f(n) = 100logb a = log10 1 = 0 Since f(n) = Θ(1) = Θ(n0), Case 2 of the Master Theorem applies. The solution, therefore, is Θ(log n).Therefore, the correct tight asymptotic bound for T(n) in 1, 2, 3 where T(1)=c is Θ(n log n).
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Let's use the Master Theorem to find the asymptotic tight bound for each of the three recurrence relations given in the problem statement.
1. T(n) = 10T(n/10) + 100n Here, a = 10, b = 10, and f(n) = 100n. We can calculate the value of logb a as follows: log10 10 = 1 Since f(n) = Θ(n1), we can apply Case 2 of the Master Theorem and get: T(n) = Θ(n log n)Therefore, the correct tight asymptotic bound for T(n) in the first case is Θ(n log n).
2. T(n) = T(n/10) + 100n Here, a = 1, b = 10, and f(n) = 100n. We can calculate the value of logb a as follows: log10 1 = 0 Since f(n) = Θ(n1), we can apply Case 1 of the Master Theorem and get: T(n) = Θ(n)Therefore, the correct tight asymptotic bound for T(n) in the second case is Θ(n).
3. T(n) = T(n/10) + 100 Here, a = 1, b = 10, and f(n) = 100. We can calculate the value of logb a as follows: log10 1 = 0 Since f(n) = Θ(1), we can apply Case 2 of the Master Theorem and get: T(n) = Θ(log n) Therefore, the correct tight asymptotic bound for T(n) in the third case is Θ(log n). Hence, the three asymptotic tight bounds for T(n) are Θ(n log n), Θ(n), and Θ(log n), respectively.
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Multi-Stage Kinematic Problems of the car as it slammed on the brakes? In some problems, there will be multiple phases where the moving object moves with different speeds or accelerations. When motion is broken up in segments like this, you need a set of kine- matic equations for each segment. (b) What is the total amount of time it takes all of this to happen? 1. According to NASA, a space shuttle launches from rest and can achieve a speed of 8,000 m/s in 8.5 minutes. How far does it ascend during this time? 3. It takes 2.9 s for a Ferrari to accelerate from 0 miles per hour to 60 miles per hour. (a) How much time does it take for the Ferrari to accelerate to 100 miles per hour from rest? (b) What is the total distance the Ferrari has traveled at the instant it is moving at 100 miles per hour? 2. A car travels at a steady 50 m/s for 100.0 m. 4. A bullet train leaves a train station and acceler- Upon seeing an obstacle, the car slams on the ates with 1 g for 60 seconds. Once up to speed brakes and decelerates, it moves with constant speed for another 60 sec- (a) What is the magnitude of the acceleration trip?
Multi-stage kinematic problems have to be broken up into segments, and a set of kinematic equations must be used for each section. The total time depends on the respective stages.
In kinematic problems, multi-stage issues have to be solved, where the motion is separated into multiple parts with different speeds or accelerations. This means that each segment must be treated as a separate problem and solved with a set of kinematic equations.
For example, a Ferrari accelerates from 0 miles per hour to 60 miles per hour in 2.9 seconds, and we have to determine the time it takes to reach 100 miles per hour from rest, and the distance it covers when it is moving at 100 miles per hour. We can solve this by dividing the motion into two sections and applying the kinematic equations to each stage. The total time depends on the respective stages, and each stage is solved separately.
Therefore, multi-stage kinematic problems have to be broken up into segments, and a set of kinematic equations must be used for each section.
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which products rely on the ability of ionic compounds to conduct electricity? check all that apply.
1. papers
2. cell phones
3. soaps
4. glazed pottery
5. remote control toys
A boy starts from point A and walks 3 meters toward the south, then turns around and walks 7 meters toward the north. What is the boys placement from point A?
Answer:
-3m+7m = 4m
Explanation:
As he walks south, he is going down 3m (-3m). Then he walks up 7m (+7m).
You subtract the final position from the initial position to get displacement.
7m - 3m = 4m
Find the volume of this figure.
4 dm
2 dm
3 dm
Answer:
4×2×3= 24dm^3
.......................
a 1000 kg car is driving around a 100 m radius curve. the coefficient of static friction is 0.80 and the coefficient of kinetic friction is 0.60. determine the maximum speed that thee car can travel without slipping.
The maximum speed that the car can travel without slipping is 28 m/s.
What is the maximum speed of the car?For a car driving around a circular path not skid at maximum speed, then the centripetal force of the car must be equal to the static frictional force.
Fc = Fs
ma = μmg
a = μg
v²/r = μg
v² = rμg
v = √rμg
where;
v is the maximum speed of the carμ is the coefficient of static frictiong is acceleration due to gravityr is the radius of the circular pathv = √(100 x 0.8 x 9.8)
v = 28 m/s.
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Avery hurls the 0.527-kg football to Tyrone the ball reaches Tyrone with a speed of 19.4 m/s Tyrone catches the pass bringing it to a dead stop in 0.010 seconds before heading up field what is the force applied to the ball ?
From Newton's second law of motion, the magnitude of the force applied is 1022.4 N
What is Force ?Force can simply be defined as a pull or push. It is the product of mass and acceleration and it is a vector quantity.
Given that Avery hurls the 0.527-kg football to Tyrone the ball reaches Tyrone with a speed of 19.4 m/s Tyrone catches the pass bringing it to a dead stop in 0.010 seconds before heading up field.
The force F applied to the ball = mass × acceleration. That is,
F = ma
Where
m = 0.527 kgv = 19.4 m/st = 0.01 sF = ?a = ?Acceleration a = v/t
a = 19.4/0.01
a = 1940 m/s²
Force F = ma
F = 0.527 × 1940
F = 1022.4 N
Therefore, the force applied to the ball is 1022.4 N
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