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Verdich [7]
3 years ago
15

2. A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? S

how all your work, including the equation used, given and unknown quantities, and any algebra required. Make sure your answer has the correct number of significant figures.
Physics
1 answer:
artcher [175]3 years ago
7 0
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.

From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T

Hence T=225/30=7.5

It took 7.5 seconds


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In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is
meriva

The spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

<h3>What is double slit experiment?</h3>

The double-slit experiment serves as a proof in current physics that both light and matter may exhibit properties of classically defined waves and particles. It also illustrates the inherently probabilistic nature of quantum mechanical events. Thomas Young carried out the first experiment of this kind employing light in 1802, illustrating how light behaves like a wave. It was formerly believed that light was made up of either waves or particles. About a century later, with the advent of modern physics, it was discovered that light may in fact exhibit behaviour like that of both waves and particles. The identical behaviour of electrons was first shown by Davisson and Germer in 1927, and it was later extended to atoms and molecules.

The separation between the slits, d = 0.05mm = 5×10⁻⁵ m

The distance from the slits to a screen, D = 2.5 m

Let x is the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits,

λ = 600nm = 6× 10⁻⁷ m  

We know that the bright fringe is given by :

y = nλD/d

So, the spacing between the first-order and second-order bright fringes is :

x = 2λD/d - λD/d

x =  λD/d

x = 6 × 10⁻⁷ × 2.5/5 × 10⁻⁵

x = 0.03 m

or

x = 3 cm

So, the spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

to learn more about double slit experiment go to -

brainly.com/question/28108126

#SPJ4

8 0
1 year ago
Can someone help me in this one :)
djyliett [7]
I believe it is “runs on parallel circuits”! my bad if incorrect
3 0
3 years ago
Water (rhoH20 = 1000.0 kg/m3 ) flows through a garden hose that goes up a step 20.0 cm high. The cross-sectional area of the hos
Soloha48 [4]

Answer:

 P₂ = 138.88 10³ Pa

Explanation:

This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation

Let's start by looking for the top speed

        Q = A₁ v₁ = A₂ v₂

We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)

         v₂ = A₁ / A₂ v₁

They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)

         v₂ = 2  1.20

         v₂ = 2.40 m / s

Now let's write the Bernoulli equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

Let's clear the pressure at point 2

       P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)

we put our reference system at the lowest point

        y₁ - y₂ = -20 cm

Let's calculate

       P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)

       P₂ = 143 103 - 2,160 103 - 1,960 103

       P₂ = 138.88 10³ Pa

3 0
3 years ago
GO TO MY LATEST QUESTION AND ANSWER <br> I WILL BRAINLIST AND GIVE 35 POINTS TO THE CORRECT ANSWER
REY [17]

y e s s s s i r r r r r r

8 0
3 years ago
Graphing: Michelle climbs a tree and drops the toy car once she has reached the top. Please create an Energy-Time graph to show
Vlada [557]

Answer:

Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.

Explanation:

  1. The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
  2. As the car falls to the ground, its PE converts to KE.
  3. When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.

<h3>PE of the car over time</h3>

The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.

Assume that air resistance on the car is negligible. The height h of the car at time t could be found with the equation:

\displaystyle h = -\frac{1}{2}\, g\, t^2 + h_0 \, (\text{Initial height}),

where

  • g \approx \rm 9.81\; m \cdot s^{-2} near the surface of the earth, and
  • h_0 is the initial height of the car.

On the other hand, \displaystyle \text{GPE} = m \, g \, h = -\frac{m \cdot g^2}{2}\, t^2 + \underbrace{m \cdot g \cdot h_0}_{\text{Initial GPE}}.

In other words, plotting the gravitational PE of the car against time would give a parabola. Since \displaystyle -\frac{m \cdot g^2}{2} < 0 (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at t = 0 the initial GPE is positive, the y-intercept of this parabola should also be positive.

<h3>KE of the car over time</h3>

Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.

To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's m \cdot g \cdot h_0. If there's no air resistance, the value of ME would stay at

Subtract PE from ME to obtain an expression for KE:

\begin{aligned} \text{KE} &= \text{ME} - \text{PE} \cr &= m \cdot g \cdot h_0 - \left(-\frac{m \cdot g^2}{2}\, t^2 + m \cdot g \cdot h_0\right)\cr &= \frac{m \cdot g^2}{2}\, t^2\end{aligned}.

That's also a parabola when plotted against t. Note that since the quadratic coefficient \displaystyle \frac{m \cdot g^2}{2} is positive, the parabola shall open upwards.

7 0
4 years ago
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