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3 years ago

A 1000 kg car drives off of u 30m cliff of

Physics

1 answer:

Inessa05 [86]3 years ago

5 0

The mechanical energy of the car is $1.54\cdot 10^6 J$

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (GPE) and its kinetic energy (KE):

E = GPE + KE

The GPE of the car is given by:

$GPE = mgh$

where

m = 1000 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

h = 30 m is the height of the car above the ground

Substituting,

$GPE=(1000)(9.8)(30)=2.94\cdot 10^5 J$

The kinetic energy of the car is given by:

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

where

m = 1000 kg

v = 50 m/s is the speed of the car

Substituting,

$KE=\frac{1}{2}(1000)(50)^2=1.25\cdot 10^6 J$

Therefore, the mechanical energy is

$E=GPE+KE=2.94\cdot 10^5 + 1.25\cdot 10^6 = 1.54\cdot 10^6 J$

Learn more about potential energy and kinetic energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts

prohojiy [21]

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

$N=N_0e^{-\lambda t }$

$.70446 =e^{-\lambda t }$

- λ t = ln .70446 = - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

7 0

3 years ago

Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly

Hunter-Best [27]

Answer:

Average velocity

$v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}$

Average speed,

$S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}$

Explanation:

(a)Average velocity

We have to find the average velocity. We know that velocity is defined as the rate of change of displacement with respect to time.

To find the average velocity we have to find the total displacement.

since displacement along east direction is 50m

and displacement along west=40m

so total displacement,

$d=50m-40m\\d=10m$

total time,

$t=28 s+42 s\\t=70 s$

therefore, average velocity

$v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}$

(b)Average Speed:

Average speed is defined as the ratio of total distance to the total time

it means

Average speed= total distance/total time

here total distance,

$D= 50m+40m\\D=90m$

and total time,

$t= 28s+40s\\t=70s$

therefore,

Average speed,

$S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}$

8 0

3 years ago

43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a

olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

$P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J$

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J$

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

$\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N$

The frictional force that acts on the sliding bear is 378.47166 N

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3 years ago

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Answer:

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Answer:

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3 years ago