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2 years ago

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The graph represents the change in that occurs when food is cooked over a charcoal grill. Which statement correctly explains the

graph?
A. The reactants are unlit charcoal that has already released its energy, and the products are charcoal that has already burned.
B. The reactants are charcoal that has already burned and released its energy, and the products are unlit charcoal.
C. The reactants are unlit charcoal, and the products are charcoal that has already burned and released its energy.
D. The reactants are charcoal that has already burned, and the products are unlit charcoal that has already released its energy.

1 answer:

MArishka [77]2 years ago

6 0

<u>O</u><u>p</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>C</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>

<h3 /><h3><em>S</em><em>m</em><em>a</em><em>l</em><em>l</em><em> </em><em>Explanation</em><em>:</em><em>-</em></h3>

The reactants are charcoal that is unlit + oxygen and the products are the burnt charcoal + energy.

(Explanation with formula and reason attached. Check it.)

$> Benjemin360$

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The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

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Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It is neccesary calculate the force:

$F=(m+m)*g*sin\theta$

Here, g = gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

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3 years ago

An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the

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To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

$m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2$

Through the aforementioned formula we will have to

$v_f^2-v_i^2 = 2ax$

The particulate part of the rest, so the final speed would be

$v_f^2 = 2gx$

$v_f=\sqrt{2(9.8)(3.1)}$

$v_f = 7.79m/s$

Now from Newton's second law we know that

$F = ma$

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

$F = m\frac{dv}{dt}$

Replacing we have that,

$F = (89)\frac{7.79}{0.5}$

$F = 1386.62N$

Therefore the force that the water exert on the man is 1386.62

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3 years ago

The above graph shows the speed of a car over time. During which time period was the car stopped?

rjkz [21]

The answer is C, as there is not increase or decrease in speed during that time frame.

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3 years ago

The format specifier ________ is a placeholder for an int value. %d %n %int %s

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%d is a format specifier that is a placeholder for an int value. It tells the compiler that we want to print an integer value that is present in variable a. In this way there are several format specifiers in c.

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3 years ago

When the acceleration of a mass on a spring is zero, the velocity is at a

Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

$F=kx$

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

$F=ma$

where

m is the mass

a is the acceleration

So we can write the equation as

$ma=kx$ (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const.$ (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a maximum

When the velocity of a mass on a spring is zero, the acceleration is at a maximum

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3 years ago