The distance travelled by the car after the brake was applied is 125 m.
What is the distance travelled by the car?
The distance travelled by the car is calculated by applying the following kinematic equation as shown below.
Mathematically, the distance travelled by the car at the given average speed is calculated as;
s = [ ( v + u ) / 2 ] t
where;
v is the final velocity of the car = 0u is the initial velocity of the car = 90 km/hr = 25 m/st is the time of motion of the car = 10 sThe distance travelled by the car is calculated as;
s = [ ( 0 + 25 m/s ) / 2 ] ( 10 s )
s = 125 m
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A soccer player runs at 10 m/s and runs into a 80 kg referee standing on the field causing the referee to fly forward at 6.0 m/s. Assume the soccer player comes to complete stop after impact. What would be the mass of the soccer player be?
The mass of soccer player will be 120kg. Writing down 1/2 m1 will help us remember that KE = 1/2 mv2 (v1i) 2 + 1/2 m2(vi) (vi) 2 = 1/2 m1(v1f) (v1f) 2 + 1/2 m2 (v2f) (v2f) 2.
How to solve ?According to Newton's third law,
Given that a soccer player runs at 10m/s and plows into a 80kg referee standing on the field causing the referee to fly forward at 56m/s. Let
mass of the player = ?
initial velocity of the player = 10m/s
Final velocity of the player
mass of the referee = 80 kg
Initial velocity of the referee = 0 ( since he is at rest )
Final velocity of the referee = 6 m/s
Since it is a perfectly elastic collision,
U1 =V2 - V1 ........ (1)
M1U1 = M2V2 + M1V1........ (2)
Substitute all the necessary parameters into the equation (1)
10 = 6 - V1
V1 = 10 - 6
V1 = 4 m/s
Now, Substitute all the necessary parameters into the equation (2)
10M1 = (80 x 6) + (4M1)
10 M1- 4 M1 = 480
4M1 = 480
M1 = 480/4
M1 = 120kg
Therefore, the mass of soccer player will be 120kg
Collision formula: What is it?When kinetic energy (KE) and momentum (p) are conserved during a collision, the collision is said to be elastic. To put it another way, it denotes that KE0 = KEf and po = pf. Writing down 1/2 m1 will help us remember that KE = 1/2 mv2 (v1i) 2 + 1/2 m2(vi) (vi) 2 = 1/2 m1(v1f) (v1f) 2 + 1/2 m2 (v2f) (v2f) 2.
How is velocity determined?By dividing the entire distance traveled by the time it took the object to go a certain distance, you may calculate the object's initial velocity. V, d, and t are the three variables that make up the equation V = d/t.
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what is the length of the y component shown below?
The length of the y component shown is C. 2.0.
How to find the length ?We are given the angle of the vector, and the length of one of the components of the vetor. Given the angle we have, the available component is the hypotenuse. The y component that we are to find, will then be the opposite or perpendicular component.
To solve for the length of the y - component therefore, the useful operation would be the Sin function.
The length of the y - component would be:
Sin 42 ° = Opposite / Hypotenuse
Sin 42 ° = y component / Hypotenuse
y - component = Sin 42 ° x Hypotenuse
y - component = Sin 42 ° x 3
y - component = 0. 6691 x 3
y - component = 2. 0
In conclusion, the y - component is 2.0.
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Full question is:
What is the length of the y-component of the vector shown below?
A. 2.2 B. 3 c. 2.0 D. 2.7For a particular nonlinear spring, the relationship betweem the magnitude of the applied force F and the resultant displacement x from equilibrium is given by the equation F = k x^2 What is the amount of work done by stretching the spring a distace x0? A) kx0^3 B) (1/2)kx0 C) (1/2)kx0^3 D) (1/3)kx0^2 E) (1/3)kx0^3
To get the work, you have to integrate the force as a function of [tex]$x$[/tex] from zero displacement to Xo
[tex](Integral of) $\mathrm{k} \mathrm{x}^{\wedge} 2 \mathrm{dx}$ from 0 to $\mathrm{Xo}_{\mathrm{o}}=(1 / 3) \mathrm{k}\left(\mathrm{Xo}^{\wedge}\right)^{\wedge} 3$[/tex]
The work done by stretching the spring to the given distance is [tex]W=\frac{k x_0}{3}[/tex]
The given parameters:
- Applied force on the spring [tex]$=F$[/tex]
- Extension of the spring [tex]$=x_0$[/tex]
The work done by stretching the spring to the given distance is calculated as follows;
[tex]W=\frac{k x_0}{3}[/tex]
[tex]$$\begin{aligned}& W=\int_{x_a}^{x_b} F d x \\& W=\int_{x_a}^{x_b} k x^2 d x \\& W=k \int_{x_a}^{x_b} x^2 d x \\& W=k\left[\frac{x^3}{3}\right] \\& W=k\left[\frac{x_b-x_a}{3}\right] \\& W=k\left[\frac{x_0-0}{3}\right] \\& W=\frac{k x_0}{3}\end{aligned}[/tex]
Thus, the work done by stretching the spring to the given distance is
[tex]W=\frac{k x_0}{3}[/tex]
measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.
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which of the following sequences lists methods for determining distance in the correct order from nearest to farthest?
Parallax, main-sequence fitting, cepheid variables, Tully-Fisher relation, and Hubble's law are the sequence lists methods for determining distance.
Parallax is the closest method for determining distance. This method uses the principle of triangulation to measure the distance of a nearby star relative to Earth. Main-sequence fitting is another method used to measure the distance of stars. This method compares the brightness of a star to other stars of known distance and luminosity to determine its distance.
Cepheid variables are stars whose brightness varies in a predictable way over time. This method uses the period of the star's brightness to determine its intrinsic luminosity, and then its distance. The Tully-Fisher relation is a method for measuring the distance of galaxies. This method uses the rotation speed of the galaxy and its brightness to calculate its distance.
Finally, Hubble's law is the most distant method for determining distance. This law states that the farther away a galaxy is, the faster it moves away from us. This allows us to calculate the distance of galaxies by measuring their rate of recession.
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a particle moves under the influence of a conservative force. the equation for the potential energy as a function of position is given by: u(x)
A particle moves under the influence of a conservative force, The maximum x-coordinate of the particle at x = -3 m is 7 m.
Given that,
Potential energy of the particle u(x) = 5x² - 20x + 2
We know the relation between force and potential energy as,
F = -du/dx = -d/dx (5x² - 20x + 2) = -(10 x - 20 ) = 20 - 10 x at mean position x = 2.
At x = -3, amplitude is given by x = 2 - (-3) = 5
The maximum x co-ordinate is given as x = 5 + 2 = 7 m
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Question
A rabbit moves a distance of 85 meters in 13 seconds.
What is the rabbits average speed?
7.26 m/s
0.15 m/s²
6.54 m/s
6.54 m/s²
Important Formula:
[tex]s=\dfrac{d}{t}[/tex]
__________________________________________________________
[tex]d=85m[/tex] (measured in meters)
[tex]t=13s[/tex] (measured in seconds)
[tex]s=?[/tex] (measured in meters per second; m/s)
__________________________________________________________
[tex]s=\dfrac{d}{t}[/tex]
[tex]s=\dfrac{85}{13}[/tex]
__________________________________________________________
[tex]\fbox{6.54 m/s},\fbox{Option C}[/tex]
based on the graphic, in what part of the electromagnetic spectrum does vegetation have the strongest response?
Between 400 nm and 700 nm of the electromagnetic spectrum, vegetation has the strongest response.
The electromagnetic spectrum travels in waves and spans an extensive spectrum from very long radio waves to very brief gamma rays. The human eye can simplest come across only a small portion of this spectrum called visible light.
In order from maximum to lowest power, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible mild, infrared radiation, and radio waves.
In a tumbler, the purple mild travels the fastest, and the violet light travels the slowest of all seven hues. Velocity and wavelength are without delay proportional.
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A large cannon is mounted on a cart with frictionless wheels that is initially at rest on a horizontal surface. The cannon fires a large cannonball to the right with a speed v_b which is then caught by a trap firmly attached to the cart. What is the final speed of the cannoncart-cannonball system? V > v_b, to the left
The final velocity of the cannonball-cart system is less than the initial velocity of the cannonball and the system will move in the direction of the cannonball ( V < v_b ).
What is the law of conservation of linear momentum?
The law of conservation of linear momentum states that the sum of the initial momentum is equal to sum of the final momentum, provided that the system is Isolated.
Mathematically, the law of conservation of linear momentum is given as;
Pi = Pf
where;
Pi is the sum of the initial momentumPf is the sum of the final momentumm₁u₁ + m₂u₂ = v ( m₁ + m₂ )
where;
m₁ is the mass cannon ballu₁ is the initial velocity of the cannon ballm₂ is the mass of the cartu₂ is the initial velocity of the cartv is the final velocity of the cannoncart-systemsince the cart is initially at rest, the initial velocity of the cart = 0
m₁u₁ + 0 = v ( m₁ + m₂ )
m₁u₁ = v ( m₁ + m₂ )
v = ( m₁u₁ ) / ( m₁ + m₂ )
given initial velocity of the cannon ball = v_b
The final velocity of the cannonball-cart system is calculated as follows;
V = ( m₁v_b ) / ( m₁ + m₂ )
Hence, V < v_b
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What is the Force of Gravity acting on the object in the diagram?
The force of gravity acting on the 23 Kg object in the diagram is 225.4 N
How do I determine the force of gravity?The force of gravity is defined as follow:
Force of gravity (F) = mass (m) × acceleration due to gravity (g)
Using the above formula, we can easily obtain the force gravity acting on the 23 Kg object. This is shown below:
Mass (m) = 23 KilogramsAcceleration due to gravity (g) = 9.8 m/s²Force of gravity (F) =?Force of gravity (F) = mass (m) × acceleration due to gravity (g)
Force of gravity = 23 Kg × 9.8 m/s²
Force of gravity = 225.4 N
Thus, from the calculation made above, we can conclude that the gravitaional force is 225.4 N
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if an object producing sound is moving away from you, you would observe a wavelength than an object moving toward you. group of answer choices
If an object producing sound is moving away from us, the wavelength of the sound heard is longer than the actual wavelength. The conclusion is from the concept of Doppler effect.
What is the Doppler effect?The Doppler's effect is a phenomenon when the source of a wave and an observer move relative to each other, the frequency heard is not the same with the actual frequency.
The equation of the Doppler effect is
f₀ = [(v ± v₀)/(v ± vs)] × fs
Where
f₀ = observer frequency of soundv = speed of sound waves (340 m/s)v₀ = observer velocityvs = source velocityfs = actual frequency of sound wavesNote:
v₀ (+) if the observer moves closer to the sound source.vs (+) if the sound source moves away from the observer.When an object producing sound is moving away from us, the frequency of the sound we heard changed.
Let's say we are at rest, it means v₀ = 0. The sound source is moving away makes vs (+).
With the Doppler's effect, we get
f₀ = [(v+0) / (v+vs)] × fs
f₀/fs = v/(v+vs)
v < v+vs
f₀ < fs
The frequency of sound we heard is lower that the actual frequency.
The wavelength is inversely proportional to the frequency. It is described in the equation:
λ = c/f
It means that the lower the frequency, the longer the wavelength.
Hence, the phenomenon which the wavelength of the sound we heard is longer than the actual wavelength when the sound source is moving away from us is called the Doppler's effect.
Here is the group of answer choices:
(a) Band width
(b) Doppler's effect
(c) Sound refraction
(d) Vibrations
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(1) A table tennis ball is dropped
onto the floor from a height of
4m and it rebounds to a height of
3m. If the time of contact with the floor is 0.01s, what is the magnitude and direction of the acceleration during the contact.
Answer:
Here, h1=4.00m,h92)=3.00m,Delta t =0.01 s.Letv1 be the velocity of the ball (actind downwards) just before striking the floor an dv20 be the velocity of the ball (acting upwaeds) ust after striking the floor. Then, change in velocity of the ball in time Δtv2−(−v1)v2+v1
:. acceleration, a=v2+v1Δt ..(i)
When body falls from height h1,
then u=o,v1,a=gandS=h1
As, v21=u2+2aS,
:. v_(1) ^(2) =0 + g h_(10 or v12–√gh1
Taking motion of the ball after striking the floor, then u=v2,v=0,a=−g,S=h2
As, v2=u22as,∴v21=0+2gh1orv_(1) =sqrt 32 g h_(1)Tak∈gmotionoftheballa>erstrik∈gthe⌊,⌋thenu=v_(2), v=0, a=- g, S=h_(2)As,v2=u2+2aS,wehave0 = v_(2)^(2) +2 (-g) h_92) otr v2=2–√gh2
Putting values in (i) we get,
a=2–√gh2+2–√gh1Δt)
a=2–√×9.8×3+2–√×9.8×40.01
= 1652m/s2.
Explanation:
