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max2010maxim [7]
3 years ago
9

5g of ammonium nitrate was dissolved in 60g of water in an insulated container. The temperature at the start of the reaction was

23.0°C and at the end it was 19.0°C. Calculate the energy absorbed by the reaction.
Physics
1 answer:
Minchanka [31]3 years ago
5 0

Answer: The energy absorbed by the reaction from the water is 996 Joules.

Explanation:

Energy absorbed by the reaction or energy lost by the water to the reaction,Q.

Mass of the the reaction  ,m = 60 g

Specific heat of water = c = 4.15 J\g ^oC

Change is temperature=\Delta T=19^oC-23^oC=-4^oC

Q=mc\Delta T=60 g\times 4.15 J\g ^oC\times (-4^oC)=-996 Joules

Negative sigh indicates that energy was given by the water to the reaction.

The energy absorbed by the reaction from the water is 996 Joules.

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What remains constant during the trajectory of an object
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3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.38 km. If the airplane rounds half
Andrej [43]

Answer:

a) 6076 m

b) 43.33 m/s

c) 68 m/s

Explanation:

(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius

s = 2r = 2* 3.38km = 6.76 km or 6760 m

(b) The average velocity would be displacement over unit of time

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(c) The length of the chord it's swept in 156s is half of the circle perimeter

c = πr = π3.38 = 10.62 km or 10620 m

The airplane average speed is its chord length over a unit of time

c / t = c / 156 = 68 m/s

4 0
3 years ago
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