Explain What determines the amount of<br> chemical energy a substance has?

Question 3 - If Julia starts out at rest, and in 13 s speeds up to 75 m/s, what is her acceleration?

Hitman42 [59]

Answer:

D

Explanation:

7 0

3 years ago

An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers

ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E where E is the value of the field and N e the charge Q

M g = N e E and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons

7 0

3 years ago

When a driver presses the brake pedal, his car stops with an acceleration of -0.5 m/s². How far will the car travel while coming

Katyanochek1 [597]

Answer:

64 m

Explanation:

Using the following symbols

x: distance

v: velocity

a: constant acceleration

t: time

v₀: initial velocity

x₀: initial position

The equations of motion for a constant acceleration are given by:

(1) x = 0.5at²+v₀t+x₀

(2) v = at+v₀

From equation (2) you can calculate the time t it takes the car to come to a complete stop.

(3) t = (v-v₀)/a

Now you plug equation (3) in equation(1):

(4) x = 0.5a((v-v₀)/a)²+v₀((v-v₀)/a)+x₀

In equation (4) the position x is the only unknown.

8 0

4 years ago

For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago

Can you answer my question please go to my profile

ad-work [718]

Answer:

what do u need help on

I went to ur profile and I don't know what I'm looking for

8 0

3 years ago

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Explain What determines the amount of chemical energy a substance has?