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adelina 88 [10]
3 years ago
10

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50

m/s2. if the car comes to a stop in a distance of 30.0 m, what was the car's original speed?
Physics
1 answer:
sergij07 [2.7K]3 years ago
7 0
To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving. 

First, we manipulate the one of the kinematic equations
 
v^2 = v0^2 + 2 (a) (x)  where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

<span>we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s 
</span>
Therefore, before stopping the car the original speed of the car would be 14.49 m/s
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A proton, an electron, and a helium nucleus all move at speed v . Rank their de Broglie wavelengths from largest to smallest.
Varvara68 [4.7K]

Ranking of de Broglie wavelengths from largest to smallest is electron > proton > helium

  • De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
  • From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship

De Broglie's relationship is given by \lambda=\frac{h}{mv}      .....(1) , where λ  is known as de Broglie wavelength and m is mass , v is velocity , h = Plank’s constant.

From equation (1) wavelength and mass has an inverse relation .

Mass of helium is 4 times the mass of the proton and proton has a greater mass than electron.

According to equation (1) , less the mass higher will be the wavelength

Hence electron having less mass have higher wavelength and then proton and then helium having large mass will have less wavelength .

Thus, order should be electron > proton > helium .

Learn about de brogile wavelength more here :

brainly.com/question/16595523

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8 0
1 year ago
PLEASE HELPP I NEED TO PASS THIS !!
goldfiish [28.3K]
Option D is correct.
3 0
2 years ago
If air temperature increased how would it effect precipitation
Oduvanchick [21]

Decrease Because Water Vapor Would Condense More Slowly

6 0
3 years ago
If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit
Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) =  (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 =  (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 =   (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0
3 years ago
The earth has a radius of 6.38 × 106 m and turns on its axis once every 23.9 h.
Vladimir [108]

Answer:

a) V = 465.9 m/s

b) θ = 70.529°

Explanation:

Let's first calculate angular velocity of earth:

\omega=\frac{2\pi}{23.9h}*1h/3600s

Velocity of a person on Ecuador will be:

V_E = \omega*R

V_E = 465.9 m/s

For part b, since angular velocity is the same:

\frac{\omega*R}{3}=\omega*(R*cos\theta )

Solving for θ:

\theta=acos(1/3)

\theta=70.529\°

8 0
3 years ago
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