40 meters divided by 5

A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

The distance the car traveled is 21.45 m

Explanation:

Motion With Constant Acceleration

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

6 0

3 years ago

Assertion – Reason

Norma-Jean [14]

u have not displayed the question fella but i can guess the answer

Explanation:

the answer is (i)

because the current in a circuit is always directly proportional to its voltage thus making it to make a straight line in its graph

3 0

2 years ago

A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil

o-na [289]

Answer:

a) U = 735 J , b) U = 125.7 J , c) U = 0 J

Explanation:

The gravitational power energy is

U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

y = 2.10 m

w = mg = 350 N

U = 350 2.10

U = 735 J

b) when the angle is 34º

y = L - L cos 34

y = L (1- cos34)

y = 2.10 (1- cos 34)

y = 0.359 m

U = 350 0.359

U = 125.7 J

c) in this case this point coincides with the reference system

y = 0

U = 0 J

4 0

3 years ago

How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00

Aleonysh [2.5K]

Answer:

 The elastic potential energy stored in the bungee cord = 20 J

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

Given: K = 10 N/m, e = 2.00 m

Substituting these values into Equation 1

E = 1/2(10)(2)²

E = 5×4

E = 20 Joules.

Therefore the elastic potential energy stored in the bungee cord = 20 J

6 0

3 years ago

A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron

Ket [755]