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ale4655 [162]
3 years ago
12

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

Physics
1 answer:
natima [27]3 years ago
5 0
The kinds of atoms in the substance and their arrangements
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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
Read 2 more answers
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