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Citrus2011 [14]
2 years ago
14

Plants use sunlight as energy to convert carbon dioxide and water into glucose and oxygen. Which best describes the reaction? Th

e law of conservation of mass indicates the same amount of carbon will be found in the reactants as in the products. The law of conservation of energy indicates the same amount of carbon will be found in the reactants as in the products. The law of conservation of mass indicates the same amount of glucose will be found in the reactants as in the products. The law of conservation of energy indicates the same amount of glucose will be found in the reactants as in the products.
Physics
2 answers:
Alina [70]2 years ago
6 0

Answer:

the correct statement is the first

The law of conservation of mass indicates the same amount of carbon will be found in the reactants as in the products.

Explanation:

The law of conservation of energy establishes that the masses are not destroyed, they can only be transformed.

Therefore the mass of carbon in the reactants (CO2 and H2O) must be in the products (glucose and oxygen)

so the correct statement is the first

The law of conservation of mass indicates the same amount of carbon will be found in the reactants as in the products.

Nikitich [7]2 years ago
3 0

Answer:

A

Explanation:

Just took the test

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The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang
STatiana [176]

the correct answers are 54N and -1,2m/s^2

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2 years ago
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On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the
Sphinxa [80]

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

d=\frac{L}{12} (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

d'=(1-\frac{1}{5})d=\frac{4}{5}d

And the number of trams will become

12+n

So eq.(1) will become

\frac{4}{5}d=\frac{L}{n+12} (2)

And substituting eq.(1) into eq.(2), we find:

\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3

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4 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

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2 years ago
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A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the
egoroff_w [7]

The acceleration of the jet is 28.1 m/s^2

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

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