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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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There are four springs stretched by the same mass. Spring A stretches 25 cm. Spring B stretches 10 cm. Spring C stretches 100 cm
saul85 [17]

Answer:

spring D stretches the weakest

6 0
2 years ago
An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t
docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

y_1(t)=h+u_1 t + \frac{1}{2}gt^2

where:

h = 50 m is the initial height

u_1=0 is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

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

So, its vertical position can be rewritten as

y_1(t)=50-4.9t^2

The position of object 2 instead can be written as

y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2

where

u_2 sin \theta is the initial vertical velocity, where

u_2 = 50 m/s is the initial velocity

\theta=30^{\circ} is the angle of projection

Substituting, we get:

y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2

The two objects collide when their vertical position is the same, so:

y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2

And solving for t, we find:

50=25t\\t= 2 s

Note that this means that the two object at t = 2 s have the  same vertical position: however, this is not true for the horizontal position.

B)

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m

We can verify that at the same time, the vertical position of object 1 is the same:

y_1(t)=50-4.9t^2=50-4.9(2.0)^2=30.4 m

This means that the two objects have the same vertical position at 30.4 m.

However, in reality, the two objects do not collide. In fact, object 1 is moving in the horizontal direction with constant velocity

v_{1x}=25 m/s

So its horizontal position at t = 2.0 s is

x_1(2.0)=v_{1x}t=(25)(2.0)=50 m

While object 2 is moving in the horizontal plane with velocity

v_{2x}=u_2 cos \theta=(50)(cos 30^{\circ})=43.3 m/s

So its horizontal position  at t = 2.0 s is

x_2(2.0)=v_{2x}t=(43.3)(2.0)=86.6 m

So in reality, the two objects do not collide, if they start from the same x-position.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
Two objects collide. Object A has a momentum of 49 kg.m/s, and Object B has a momentum of
Natalka [10]

Answer:

O 47 kg. m/s if they were initially headed opposite direction

O 145 kg. m/s if they were initially headed the same direction

The other two are possibilities as well if velocities are initially at an angle

Explanation:

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kozerog [31]

Answer:

Some work input is lost to friction

Explanation:

The efficiency of a machine is defined as:

\eta = \frac{W_{out}}{W_{in}} (1)

where

W_{out} is the work output

W_{in} is the work input

Due to the law of conservation of energy, the work output can never be larger than the work input (because energy cannot be created). Moreover, in real machines part of the work input is lost due to the presence of frictions: as a result, part of the energy in input is converted into thermal energy or other forms of energy, and so the work output is smaller than the work input, and so the ratio (1) becomes less than 1, and so the efficiency is less than 100%.

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False, while holding the book it means there is no work done on the book.
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