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2 years ago

Why is figure 5 an unhelpful visualization tool for this data set? Please help!

Physics

1 answer:

Paraphin [41]2 years ago

8 0

Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

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The molecule that traps the sun's energy is ?.

kakasveta [241]

The molecule that trap's the sun's energy is chlorophyll.

7 0

3 years ago

Which substance is used to remove rust from metal?

Vlad [161]

Answer : The correct option is, (B) Hydrochloric acid(HCl)

Explanation :

Rust is an iron oxide, which is formed by the reaction of iron with oxygen in the presence of water.

As we know that the rust is an iron oxide (base) which is removed by an acid.

As per given options, the acid used to remove the rust from the metal is hydrochloric acid and not sulfuric acid because sulfuric acid is an oxidizing agent which oxidizes the metal more.

While the other options, ammonia and sodium hydroxide are a base.

Hence, hydrochloric acid is used to remove rust from metal.

7 0

3 years ago

Read 2 more answers

A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistan

Alexxx [7]

Answer:

a) $t_{1}=3.49s$

b) $t_{2}=2.00s$

c)Xmax=80.71m

Explanation:

a)Kinematics equation for the Stone, dropped:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=0m/s$ the stone is dropped

The ball reaches the ground, y=0, at t=t1:

$0=h-1/2*g*t_{1}^{2}$

$t_{1}=\sqrt{2h/g}=\sqrt{2*60/9.83}=3.49s$

b)Kinematics equation for the Stone, with a initial speed of 20m/s:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=-20m/s$ the stone is thrown straight down

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{2}-1/2*g*t_{2}^{2}$

$0=60-20t_{2}-1/2*9.83*t_{2}^{2}$

t2=-6.01 this solution does not have physical sense

t2=2.00

c)Kinematics equation for the Stone, with a initial speed of 20m/s with an angle of 30° above the horizontal:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=20sin(30)=10m/s$ the stone is thrown with an angle of 30° above the horizontal

The ball reaches the ground, y=0, at t=t3:

$0=h+v_{o}t_{3}-1/2*g*t_{3}^{2}$

$0=60+10t_{3}-1/2*9.83*t_{3}^{2}$

t3=-2.62 this solution does not have physical sense

t3=4.66

the movement in x:

v=constant=20cos(30)m/s

x(t)=v*t

Xmax=v*t3=20cos(30)*4.66=80.71m

3 0

3 years ago

An object accelerates from rest, with a constant acceleration of 14.5 m / s, what will its velocity be after 92 s?

aniked [119]

Answer:

If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s? v0 = 0 v= ? a = 5.4 m/s2. ∆x = t = 28 s. 3.

Explanation:

6 0

3 years ago

Standing on the roof of a (42.0+A) m tall building, you throw a ball straight up with an initial speed of (14.5+B) m/s. If the b

RSB [31]

First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:

2as = v² - u²

Because the final velocity v is 0 in such cases

s = -u²/2a; because both u and a are downwards, the negative sign cancels

s = 14.5² / 2*9.81
s = 10.72 meters

Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m

We will use the formula
s = ut + 0.5at²

to find the time taken with the initial velocity u = 0.

55.72 = 0.5 * 9.81 * t²

t = 3.37 seconds

4 0

3 years ago