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anzhelika [568]
3 years ago
8

If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step

Physics
1 answer:
azamat3 years ago
8 0

Answer:

Explanation:

v_f^2 = v_i^2-2a(x_f-x_i)

Subtract both sides by v_i^2:

- v_i^2+v_f^2 = -2a(x_f-x_i)

Divide both sides by -2*a:

\frac{v_i^2 - v_f^2}{2a} =x_f-x_i

Add both sides by x_i:

x_i+\frac{v_i^2 - v_f^2}{2a} =x_f

Subtract both sides by \frac{v_i^2 - v_f^2}{2a}:

x_i=x_f-\frac{v_i^2 - v_f^2}{2a}

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The frequency of this wave is 3
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James took two pea plants, placing one in a dark closet and the other on a sunny window sill. Both are located in air-conditione
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The constant is the temperature of the air that the plants get.

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A 10 kg blue cart moving to the right at 25 m/s collides with a 17 kg red cart moving in the opposite direction at 16 m/s. If, a
scoray [572]

Answer:

24.8m/s

Explanation:

Given data

m1= 10kg

u1=25m/s

m2=17kg

u2=16m/s

v1=10m/s

v2=??

Applying the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

10*25+17*16=10*10+17*v2

250+272=100+17v2

522=100+17v2

522-100=17v2

422=17v2

Divide both sides by 17

v2= 422/17

v2= 24.8 m/s

Hence the velocity of the red cart is 24.8m/s in the opposite direction of the blue cart

3 0
2 years ago
Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor
rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

e_{1}=\frac{E}{4\pi r^{2}}

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}

Hence we sense it as 4 times fainter than the nearer star.

5 0
3 years ago
You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr
Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

v_r = 27 - 10 = 17 m/s

relative deceleration of car

a_r = -7 m/s^2

now the distance before they stop with respect to each other is given by

v_f^2 - v_i^2 = 2 a d

0 - 17^2 = 2 *(-7)*d

d = 20.6 m

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

v_f^2 - v_i^2 = 2 a s

0^2  - 27^2 = 2 * (-7)* s

s = 52.1 m

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

v_f - v_i = at

0 - 27 = (-7) * t

t = 3.86 s

now the distance covered by truck in same time

d = 3.86 * 10 = 38.6 m

now after the car will stop its distance from the truck is

D = 25 + 38.6 - 52.1 = 11.5 m

<em>so the distance between them is 11.5 m</em>

6 0
3 years ago
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