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pickupchik [31]
3 years ago
7

3. Infer A car is travelling down a

Physics
1 answer:
Zinaida [17]3 years ago
8 0

Answer:

because the car is broken.

Explanation:

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sveticcg [70]


A.) is chemical, B.) is physical, C.) is physical, D.) is chemical, E.) is physical, F.) is physical, G.) is physical, and H.) is chemical.

5 0
3 years ago
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Thermal energy depends on an object’s
oee [108]

The correct answer is D. I alread took this test.

4 0
3 years ago
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When the voltage across a steady resistance is doubled, the current?
natima [27]

I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.

<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know V and I use formula P = IV: P = IV = (100mA)(10V) = 1 W.</span>

The next question is what I'm not sure about:

Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).

What I did initially was: P = IV = (100mA)(2V) = 2 W

But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."

So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.

P = IV = (200mA)(2V) = 4 W

8 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the
lilavasa [31]

Answer:

R=803k\Omega

Explanation:

We have the following information,

V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s

We apply the equation for capacitor charging the voltage across it,

V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}

Replacing values,

R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}

R=803k\Omega

3 0
3 years ago
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