A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60
what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.
Answers
Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
Related Questions
what is photosynthesis
Answers
: [tex] \implies[/tex] The Photosynthesis is the process of capturing light energy and transforming it into chemical energy. Green plants and several other organisms use light energy and convert carbon dioxide and water into glucose. In this process, oxygen is produced as a by-product
→ We also who how it's process occur
In plants and blue-green algae, the photosynthesis process takes place in chloroplasts. The chloroplast is present in all green parts of a plant – the leaves, green stems, sepals, and even in the flowers, in the form of green colour plastids. The chloroplast is found only in plant cells and is essential for photosynthesis reaction.Photosynthesis Equation
Carbon dioxide and water are the two major factors involved in the photosynthesis reaction. It’s an endothermic reaction, and the products resulting from it are oxygen and glucose. The formula is:6CO2 + 6H2O = C6H12O6 + 6O2
However, some bacteria don’t produce oxygen as a by-product of photosynthesis. They are called anoxygenic photosynthetic bacteria, and those who do it are called oxygenic photosynthetic bacteria.Importance of Photosynthesis
The photosynthesis process is very important for the survival of living beings, and to continue the food chain. It also produces oxygen, which is required for breathing.Photosynthetic Pigments
Four types of photosynthetic pigments are present in the leaves of the plants. They are: Chlorophyll a chlorophyll b xanthophylls CarotenoidsThe Factors Affecting Photosynthesis
Various factors influence/affect the photosynthesis process. These are:
Light Intensity: More the light, the more will be the rate of photosynthesis. Similarly, low light will lead to a low rate of photosynthesis.The Concentration of CO2: A higher CO2 concentration rate in a plant also accelerates the photosynthesis process. The required amount of CO2 is 300-400 PPM.Temperature: If the temperature is between the range of 25 to 35 degrees Celsius, the photosynthesis takes place effectively.Water: An essential amount of water is required for stomatal opening, and it’s a key factor in the process of photosynthesis.Pollution: The increasing rate of polluting particles in the atmosphere block the pores of somatic cells, and the intake of carbon dioxide becomes difficult.▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Hope it's helps !!
If the penny is thrown horizontally at 25 m/s from the 170 meter building, how long will it take for the penny to hit the ground?
Answers
9514 1404 393
Answer:
about 5.89 seconds
Explanation:
The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...
h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters
h(t) = 0 when ...
4.9t^2 = 170
t = √(170/4.9) ≈ 5.89
The penny will hit the ground in about 5.89 seconds.
A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resistance of the dryer?
Answers
Answer:
22Ω
Explanation:
if V ⇒ voltage
I ⇒ current
R ⇒ resistance
V = IR
220 = 10 x R
220 / 10 = R
22 = R
What is the direction of the torque produced on the crankset by the 2-kg mass attached to the pedal bar
Answers
A Torque is a twisting force, or turning moment, it is a vector quantity with both magnitude and direction e.g Turning the handle of a cork-screw clockwise and then counterclockwise will advance the screw first inward and then outward By convention, counterclockwise torques are positive and clockwise torques are negative.
The direction is perpendicular to both the radius from the axis and to the force. It is conventional to choose it in the right hand rule direction along the axis of rotation.
Counterclockwise is the positive rotation direction and clockwise is the negative direction.
Learn more at:
https://brainly.com/question/18253139
A 0,9 -kg object attached to the end of a string swings in a vertical circle (radius = 75 cm). At the top of the circle the speed of the object is 6,5 m/s. What is the magnitude of the tension in the string at this position?
Answers
The car on this ramp starts from rest. When released, it
accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?
Answers
Answer:
Explanation:
s 0.12 + 1.20(1.80) = 2.28 m from the top.
A 2 kg ball is rolling down a hill at a constant speed of 4 m/s. How much kinetic energy does the ball have?
Answers
0.5x2x16
So the answer is 16
The control center of human body is * 1/1 heart
brain
liver
We have ________ sense organs * 1/1
6
4
5
Which energy we get by burning of fuels * 1/1
Solar
Heat
Light
Sources of energy that can be used again and again will never run out * 1/1 Non renewable energy Renewable energy Running water What is called structure that build on rivers to get electricity * 1/1 Dam Check dam Tunnel Which nutrients protect us from diseases * 1/1 Carbohydrate Vitamin and Mineral Protein When is the best time of the day to water your lawn * 1/1 Early morning and late evening Afternoon Night Which of these ways to wash the car saves the most water? * 1/1 Wash it in the driveway with the garden hose Drive it into the lake Take it through a car wash that recycles water Why do we keep food and vegetables in refrigerator? * 1/1
Answers
What is a list of all the states of matter?
Answers
Answer:
3
Explanation:
state of matter are solid
liquid and
gases
Answer:
3
Explanation:
state of a matter are solid liquid and gas
A 1.1 kg ball drops vertically onto a floor, hitting with a speed of 23 m/s. It rebounds with an initial speed of 5.0 m/s. (a) What impulse acts on the ball during the contact
Answers
Hi there!
We know that:
I = Δp = m(vf - vi)
Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.
I = 1.1(5 - (-23)) = 30.8 Ns
A flywheel having constant angular acceleration requires 4.70 s to rotate through 164 rad . Its angular velocity at the end of this time is 101 rad/s . Find the angular velocity at the beginning of the 4.70 s interval. Find the angular acceleration of the flywheel.
Answers
Answer:
A) -31.2 rad/s
B) 28.1 rad/s^2
Explanation:
An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]
Answers
a.
The object's speed at 2.20 m below balcony level is 8.74 m/s
Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.
Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and
So, E = E'
U + K + f = U' + K' + f'
where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).
So,
U + K + f = U' + K' + f'
mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh
mgh = mgh' + 1/2mv² + 0.10mgh
Dividing through by m, we have
gh = gh' + 1/2v² + 0.10gh
So, gh - 0.10gh = gh' + 1/2v²
0.90gh = gh' + 1/2v²
1/2v² = 0.90gh - gh'
1/2v² = g(0.90h - h')
v² = 2g(0.90h - h')
Taking square-root of both sides, we have
v = √[2g(0.90h - h')]
where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the values of the variables into the equation, we have
v = √[2g(0.90h - h')]
v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]
v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]
v = √[2 × 9.8 m/s²(3.901 m)]
v = √[76.4596 m²/s²]
v = 8.74 m/s
So, the object's speed at 2.20 m below balcony level is 8.74 m/s
b.
Yes it does matter when we apply 10% loss before V calculations
We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.
So, yes it does matter when we apply 10% loss before V calculations
Learn more about speed of a falling object here:
https://brainly.com/question/18732868
please paraphrase this paragraph
As a conclusion we first learned what is Discrete Fourier Transform and
why is it very important in signal processing science as it helps us to convert the discrete signal from time domain into frequency domain and the inverse DFT does the opposite converting the signal from frequency domain back into time domain. We have also learned the difference between DFT and the FFT algorithm where the FFT is an efficient
algorithm created to simplify the DFT process as it is faster and more accurate. We also learned how to perform and implement this algorithm using Matlab,
Answers
Answer:
The importance of signal processing science as has been made clear. Since we now know that it help to convert discrete signal from time domain to frequency domain. Moreover, we are now aware of the difference in DF5 and FFT algorithms as well as the implementation in Matlab.
The center of mass of a 1600 kg car is midway between the wheels and 0.7 m above the ground. The wheels are 2.6 m apart. (a) What is the minimum acceleration A of the car so that the front wheels just begin to lift off the ground
Answers
Answer:
Explanation:
I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero
If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.
Call the traction force at the rear wheels F
The normal force on the front wheels will be zero, so no moment generated by the front wheels.
Summing moments about our chosen point to zero
1600(9.8)[2.6 / 2] - F[0.7] = 0
F = 291,200
this force will create an acceleration of
a = F/m
a = 291200/1600
a = 182 m/s²
which is about 18.6 times gravity acceleration
Objects 1 and 2 attract each other with a gravitational force of 179 units. If the distance separating objects 1 and 2 is changed to four times the original value (i.e., quadrupled), then the new gravitational force will be ______ units.
Answers
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant throughout the universe.
D is the distance between the 2 objects.
the distance is now quadrupled.
Fgravitynew = G*(mass1*mass2)/(4D)² =
= G*(mass1*mass2)/(16D²) =
= (G*(mass1*mass2)/D²) / 16 = Fgravity/16
the new gravitational force will be 179/16 = 11.1875 units
A car with an initial position of 10.0 m
and an initial velocity of 16.0 m/s accelerates at an average rate of 0.50 m/s2 for 4.0 s. What is the car’s position after 4.0 s?
Answers
Answer:
78
Explanation:
x=xi+vi(t)+1/2a(t)^2
x=10+16(4)+1/2(0.50)(4)^2
x=74+4
x=78 m
PLEASE HELP!
A 9kg particle is initially at rest at x=0. It is subject to a single force Fx (N) which varies with x (m) as shown in the
diagram
F
2
1
0
ББ by
x
- 1
1
-2
The kinetic energy of the particle when it is at x = 3 m is:
Answers
Hi there!
With a Force/Displacement curve, we must take the integral (area underneath the curve) to calculate the work done.
We know that:
W = ΔKE
Calculate the work by finding the area underneath the force curve from
x = 0 to 3 m:
We can use a trapezoid:
A = 1/2(3 + 2)(3) = 7.5 J
This is the amount of work done, and since the object starts from rest:
7.5J = KEf - KEi (0 J)
7.5J = KEf = 7.5 J
An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 15.0 kg, a radius of 0.400 m, and a length of 0.800 m. The mass of the end of the barrel equals a fourth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is =9.80 m/s2.
What is the translational speed f of the barrel at the bottom of the hill if released from rest at a height of 33.0 m above the bottom?
Answers
Hi there!
We can use work and energy to solve this problem.
We know that:
Ei = Ef
Ei = Potential energy = mgh
Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²
The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:
I (hollow cylinder) = mr²
I (disk) = 1/2mr²
Calculate the moment of inertias of each.
Since the mass on the base is one-fourth of its side:
x = mass of side
x + x/4 = 15
4x + x = 60
5x = 60
x = 12 kg
end mass = 3 kg
Solve for each moment of inertia:
Side: (12)(0.4²) = 1.92 Kgm²
Bottom: 1/2(3)(0.4²) = 0.24 Kgm²
Side + bottom = 2.16 Kgm²
We can now solve:
mgh = 1/2mv² + 1/2(2.16)v²/r²
(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²
4851 = 14.25v²
v = 18.45 m/s
A tennis player strikes the tennis ball with an initial velocity of 44.7 m/s horizontally. The ball is initially 1.28 m above the ground and 12.9 m from the 0.914 m tall net. Does the tennis ball make it over the net?
Answers
Hi there!
We can begin by finding the total time taken for the ball to reach the net using the equation:
dₓ = vₓt
12.9 = 44.7t
12.9/44.7 = t = 0.289 s
Now, we can use the following equation to solve for displacement in the Y direction:
d = y₀ + vit + 1/2at²
There is no initial vertical velocity, so:
d = y₀ + 1/2at²
Plug in known values:
d = 1.28 + 1/2(-9.8)(0.289²)
d = 0.87m
Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.
Suppose you lift a 17 kg k g box by a height of 1.0 m m . How much work do you do in lifting the box
Answers
Answer:
17 J
Explanation:
work done= force X distance moved
What happens when the object is placed at F? Explain
your answer.
Answers
Answer:
Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.
Explanation:
Which is the most famous book of the philosopher Alexis karpouzos? I think it is the ''Cosmology, philosophy and physics''.
Answers
Answer: Yes, the " Cosmology, philosophy and physics" is the most famous book of the philosopher, Alexis karpouzos. But and the other books are important. For example, the " The self-criticism of science", the "Universal conscilusness" and the "Non-duality".
Explanation:
14 The radius of gyration of a body about an axis &ta
distance 6 cm from its centre of mass is 10 cm.
Then, its radius of gyration about a parallel axis
through its centre of mass will be
(a) 80 cm (b) 8 cm (c) 0.8 cm (d) 0.08 cm
Answers
Correct option is B 8 cm.
Let radius of gyration for the axis not passing through center of mass be r and that for the axis passing through the center of mass be k and the distance between the two parallel axes be a.
Parallel axes theorem gives:
[tex] {mr}^{2} = m( {k}^{2} + {a}^{2} ) \\ ⇒ {r}^{2} = {k}^{2} + a {}^{2} [/tex][tex]⇒k = \sqrt{ {10}^{2} - {6}^{2} } = 8cm.[/tex]Thus, option B is the correct answer.
All circuits include
a battery, wires, and a switch.
an energy source, a resistor, and a battery.
a battery, a light bulb, and a switch.
an energy source, a load, and wires.
Answers
Answer:
a battery, wires, and a switch.
Explanation:
All circuits include?
Someone with a force of 900 N does not float in a freshwater pool. To prevent them from sinking, 20 N must be applied in an upward fashion. What is their volume and average density?
Answers
Explanation:
The buoyant force [tex]F_B[/tex] is defined as
[tex]F_B = \rho_wgV[/tex]
where [tex]\rho_w[/tex] is the density of the displaced fluid (freshwater), g is the acceleration due to gravity and V is the volume of the submerged object. In the case of freshwater, its density is [tex]997\:\text{kg/m}^3.[/tex] Since the buoyant force is 20 N, we can solve for the volume of the displaced fluid:
[tex]F_B = \rho_wgV \Rightarrow V = \dfrac{F_B}{\rho_wg}[/tex]
Plugging in the values, we get
[tex]V = \dfrac{20\:\text{N}}{(997\:\text{kg/m}^3)(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:= 2.05×10^{-3}\:\text{m}^3[/tex]
Recall that the weight of an object in terms of its density and volume is given by
[tex]W = \rho gV[/tex]
Using the value for the volume above, we can solve for the density of the object as follows:
[tex]\rho = \dfrac{W}{gV} = \dfrac{900\:\text{N}}{(9.8\:\text{m/s}^2)(2.05×10^{-3}\:\text{m}^3)}[/tex]
[tex]\:\:\:\:\:= 44,798\:\text{kg/m}^3[/tex]
Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.150 rev/srev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.
What is the radius of the circle in which Jack travels? Treat him as a point mass.
Answers
Answer:
Explanation:
At the top of the arc, 3/4 of the acceleration of gravity is use to supply the necessary centripetal acceleration.
0.75g = ω²R
R = 0.75g/ω²
R = 0.75(9.81) / (0.15 rev/s)(2π rad/rev)²
R = 8.283006...
R = 8.28 m
a man can jump 9meteres on the moon.how high can he jump on the earth.
Answers
Answer:
You can jump 1.5 feet on Earth.
Explanation:
Because the moons gravity is weaker that earth so it would be easier to jump further on the moon.
Why metals have thermoconductivity higher than ceramic?
Answers
Answer:
Thermal Conductivity Easily Transmits Heat Among Fine Ceramics
The_____ scale is called an absolute temperature scale, and its zero point is called absolute zero.
Kelvin
Fahrenheit
Celsius
Answers
The Celcius Scale is called an absolute Temperature scale,and it's zero point called absolute zero
_______________
For the ballistic missile aimed to achieve the maximum range of 9500 km, what is the maximum altitude reached in the trajectory
Answers
Explanation:
The range R of a projectile is given the equation
[tex]R = \dfrac{v_0^2}{g}\sin{2\theta}[/tex]
The maximum range is achieved when [tex]\theta = 45°[/tex] so our equation reduces to
[tex]R_{max} = \dfrac{v_0^2}{g}[/tex]
We can solve for the initial velocity [tex]v_0[/tex] as follows:
[tex]v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}[/tex]
or
[tex]v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}[/tex]
[tex]\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}[/tex]
To find the maximum altitude H reached by the missile, we can use the equation
[tex]v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy[/tex]
At its maximum height H, [tex]v_y = 0[/tex] so we can write
[tex]0 = (v_0\sin{45°})^2 - 2gH[/tex]
or
[tex]H = \dfrac{(v_0\sin{45°})^2}{2g}[/tex]
[tex]\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:\:= 2.4×10^6\:\text{m}[/tex]