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3 years ago

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How do we know that caramelizing sugar is a chemical change? A. It involves dissolving sugar in water. B. It changes the state o

f sugar from a solid to a liquid. C. It separates sugar from a sugar solution. D. It changes the sugar's color.

1 answer:

amid [387]3 years ago

8 0

Answer:

I think the correct answer is c

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What is the weight of a 48kg rock?

IrinaK [193]

Answer:

48kg

Explanation:

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3 years ago

Which shows the correct lens equation? The inverse of f equals the inverse of d Subscript o Baseline times the inverse of d Subs

musickatia [10]

Answer:

The inverse of f equals the inverse of d Subscript o Baseline plus the inverse of d Subscript I Baseline.

Explanation:

The lens equation shows the relation among focal length of the lens, image distance and object distance. It can be expressed as:

$\frac{1}{f}$ = $\frac{1}{d_{o} }$ + $\frac{1}{d_{I} }$

where: f is the focal length of the lens, $d_{o}$ is the object distance to the lens and $d_{I}$ is the image distance to the lens.

The lens equation can be used to determine the unknown value among the variables f , $d_{o}$ and $d_{o}$ .

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3 years ago

Read 2 more answers

A truck moves 100 miles to the south in 2 hours. What is the trucks velocity

ruslelena [56]

v = d/t

v = 100/2

v=50

50 mph

3 0

3 years ago

A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj

Norma-Jean [14]

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

8 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago