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lana [24]
4 years ago
9

Examples of stored energy

Physics
1 answer:
jeyben [28]4 years ago
3 0

Answer:

Biomass, petroleum, natural gas, and propane are examples of stored chemical energy. Energy is energy stored in objects by the application of a force

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Eight levels of classification from most general to most specific
Airida [17]
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5 0
3 years ago
A dynamite blast propels a heavy rock straight up with a launch velocity of 160ft/sec (about 109 mph). Write a position for the
Flura [38]
A) Using:
2as = v² - u², where v will be 0 at max height
s = -(160)² / 2 x -32.174
s = 397.8 ft

b) Using:
s = ut + 1/2 at²
256 = 160t - 16.1t²
solving for t,
t = 2.0, t = 7.9
Now, v = u + at, for both times:
v(2) = 160 - 32.174(2)
v(2) = 95.7 ft/sec on the way up

v(7.9) = 160 - 32.174(7.9)
v(7.9) = -94.3 ft/sec; 94.3 ft/sec on the way down

c) -32.174 ft/s², which is the acceleration due to gravity.

d) s = 0
0 = 160t - 1/2 x 32.174t²
t = 9.94 seconds
3 0
4 years ago
What effect does gender have on sense of smell?
Anastasy [175]

Gender for a guy would be like old smelly socks in a closet

gender for a girl would smell like flowers and perfumes

Such a big difference

7 0
3 years ago
Read 2 more answers
A child pulls on a wagon with a horizontal force of 75 N. If the wagon moves horizontally a total of 42 m in 1.5.0 min, what is
d1i1m1o1n [39]

Answer:

B) 35 W

Explanation:

Force applied by child =  75 N = F

Distance travelled by child = 42 m = d

Time traveled for is 1.5 min = 1.5×60 = 90 seconds = t

Work done by the child

W = Fdcosθ

⇒W = 75×42cos0

⇒W = 3150 Joule

Power is defined as work done per unit time

P=\frac{W}{t}\\\Rightarrow P=\frac{3150}{90}=35\ W

∴ The average power generated by the child is 35 W

4 0
4 years ago
If an object is thrown upward with an initial velocity of 128 ​ft/second, then its height after t seconds is given by the follow
IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

h = 128t-32t^2....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

\dfrac{dh}{dt}=0

From equation (I)

\dfrac{dh}{dt}=128-64t

128-64t=0

t=\dfrac{128}{64}

t=2\ sec

Now, for maximum height

Put the value of t in equation (I)

h =128\times2-32\times4

h=128\ ft

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

128t-32t^2=0

t(128-32t)=0

128=32t

t=4\ sec

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0
3 years ago
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