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2 years ago

A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed

Physics

1 answer:

malfutka [58]2 years ago

5 0

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here: brainly.com/question/25573309

#SPJ10

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Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?

liberstina [14]

Answer : The maximum concentration of silver ion is $5\times 10^{-12}m$

Solution : Given,

$K_{sp}$ for AgBr = $5\times 10^{-13}$

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

$NaBr(aq)\rightleftharpoons Na^++Br^-$

The concentration of NaBr solution is 0.1 m that means,

$[Na^+]=[Br^-]=0.1m$

The equilibrium reaction for AgBr is,

$AgBr\rightleftharpoons Ag^++Br^-$

At equilibrium s s

The expression for solubility product constant for AgBr is,

$K_{sp}=[Ag^+][Br^-]$

The concentration of $Ag^+$ = s

The concentration of $Br^-$ = 0.1 + s

Now put all the given values in $K_{sp}$ expression, we get

$5\times 10^{-13}=(s)(0.1+s)$

By rearranging the terms, we get the value of 's'

$s=5\times 10^{-12}m$

Therefore, the maximum concentration of silver ion is $5\times 10^{-12}m$ .

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2 years ago

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A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a) Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 1.5 m

Substituting

s = ut + 0.5 at²

1.5 = 0 x t + 0.5 x 9.81 xt²

t = 0.553 s

Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 200 m/s

Acceleration, a = 0 m/s²

Time, t = 0.553 s

Substituting

s = ut + 0.5 at²

s = 200 x 0.553 + 0.5 x 0 x 0.553²

s = 110.6 m

The bullet travels horizontally 110.6 m

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