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3 years ago

Think of ways you control temperature to influence chemical changes during a typical day. (Hint: cooking, art class)

1 answer:

3 0

Here are some ways that you can control the temperature to influence chemical changes during a typical day.
1. Boiling/Cooking of food such as eggs, soup, meat, vegetables, etc.
2. Baking that includes making a caramel from sugar, and turning other ingredients into some pastries.
3. Burning of wood used for cooking

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Solid barium sulfate is placed into a beaker to form a saturated solution of barium sulfate. the solution has a barium concentra

borishaifa [10]

Ksp = [Ba⁺²][SO₄⁻²]

[Ba⁺²] = [SO₄⁻²] for barium sulfate
Thus,
Ksp = (1 x 10⁻⁵)²
Ksp = 1 x 10⁻¹⁰

5 0

3 years ago

Read 2 more answers

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

Read 2 more answers

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago

A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin

emmainna [20.7K]

(1)

Cheetah speed: $v_c = 97.8 km/h=27.2 m/s$

Its position at time t is given by

$S_c (t)= v_c t$ (1)

Gazelle speed: $v_g = 78.2 km/h=21.7 m/s$

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

$S_g(t)=S_0 +v_g t$ (2)

The cheetah reaches the gazelle when $S_c=S_g$ . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle: