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bixtya [17]
2 years ago
14

Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.

Physics
1 answer:
LiRa [457]2 years ago
7 0

Answer:

         Time - taken = 2.5 s

          deceleration= -8 m/s²

Solution:

            Given:

                     speed, v = 8 m/s

                 distance, d = 20m

                     

              To Find:

                     deacceleration = ?

               

               As we know speed is defined as

                          v = d/t

                plugging in the values

                          t =  20/ 8

                          t = 2.5s

                Now from deceleration formula

                        a =  - v/ t

                        a = - 20/ 2.5

                        a = - 8 m/s²

          Thus, the time taken and acceleration is 2.5 s and -8 m/s²

          respectively.

          Learn more about deceleration here:

                brainly.com/question/13354629

                       #SPJ4  

               

                       

             

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Explanation:

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These clouds when get saturated then fall in the form of precipitation of water, snow etc. Replenishing back the water of the earth.

7 0
3 years ago
The half-life of iodine-131 is 8.04 days. It’s decay constant equals
Svetllana [295]

Answer:

Explanation:

we know that half life of an element is

T=0.693/λ

where λ is decay constant in order to find decay constant

λ=0.693/T

λ=0.693/8.04

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7 0
3 years ago
What is its maximum altitude above the ground? The answer is the maximum height above the ground
Kisachek [45]

Answer:

Maximum altitude above the ground = 1,540,224 m = 1540.2 km

Explanation:

Using the equations of motion

u = initial velocity of the projectile = 5.5 km/s = 5500 m/s

v = final velocity of the projectile at maximum height reached = 0 m/s

g = acceleration due to gravity = (GM/R²) (from the gravitational law)

g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)

g = -9.82 m/s² (minus because of the direction in which it is directed)

y = vertical distance covered by the projectile = ?

v² = u² + 2gy

0² = 5500² + 2(-9.82)(y)

19.64y = 5500²

y = 1,540,224 m = 1540.2 km

Hope this Helps!!!

3 0
3 years ago
What is special about decomposers? *
Bogdan [553]
D. They are heterotrophs that digest food internally.
5 0
3 years ago
A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

3 0
3 years ago
Read 2 more answers
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