Use newton's method to find the second and third approximation of a root of

A car increases its velocity from zero to 97 km/h in 8.4 s. What is its acceleration?

ser-zykov [4K]

Answer:

$a=\frac{v-u}{t}$

$v=$ final velocity =97km/h(change to m/s) =26.9m/s NEVER FORGET TO DO THAT BECAUSE TIME IS GIVEN IN SECONDS

$u=$ initial velocity 0 m/s

$t=$ time 8.4s

$a=\frac{26.9-0}{8.4}$

$a=$ $3.2m/2^{2}$

3 0

3 years ago

A boy drops a coin down a well that is 225 m deep. How long does it take the coin to hit the bottom of the well? please help

Nady [450]

<span>It takes 6.78 seconds for the coin to hit the bottom of the well. We can use the equation h = 0.5gt^2, where h is the height of the coin, g is the gravitational constant of 9.8m/s^2, and t is the time is takes for the coin to hit the bottom of the well. Solve for t to obtain 6.87 seconds.</span>

8 0

3 years ago

Which does energy never do?

statuscvo [17]

It never disappears I believe

3 0

3 years ago

A bucket filled with water has a mass of 23 kg and is attached to a rope, which in turn, is wound around a 0.050-m radius cylind

Cloud [144]

Answer:11.27 Nm

Explanation:

Given

mass of bucket with water is $m=23\ kg$

radius of cylinder $r=0.05\ m$

If the cylinder is not permitted to rotate

Torque is given by the product of Force and distance of force from axis of rotation

$T=F\times r$

where F=weight of bucket with water

$F=mg$

$T=mg\times r$

$T=23\times 9.8\times 0.05$

$T=11.27\ Nm$

5 0

3 years ago

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$