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3 years ago

6

I need help this is for economics, can anyone help me ?

1 answer:

PolarNik [594]3 years ago

4 0

i can <u>when do u need it by</u> working on it now!!!

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How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:

zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

$W=\Delta K$

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

$K=\frac{1}{2}mv^2$

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

$\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\ \\ \therefore K\approx532,346J \end{gathered}$

Therefore, the answer is: 532,346 J.

5 0

1 year ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago

Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and fricti

julsineya [31]

Answer:

5N

Explanation:

(25 N - 20 N = 5 N)

3 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an

Juliette [100K]

Answer: $W=1.93\times 10^9 J$

Explanation:

Given

Force $F=1.6\times 10^{6} N$

one at an angle of $13^{\circ}$ East of North and another at $13^{\circ}$ West of North

Net Force is in North Direction

$F_{net}=2F\cos 13$

Forces in horizontal direction will cancel out each other

thus Work done will be by north direction forces

$W=2F\cdot \cos 30\cdot s$

here $s=0.7 km$

$W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700$

$W=1.93\times 10^9 J$

3 0

3 years ago

Solve for real t: -25=20t-5t^2

PolarNik [594]

Answer should be <span>t=<span>−<span><span>1<span> or </span></span>t</span></span></span>=<span>5</span>

3 0

3 years ago