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sesenic [268]
3 years ago
7

A 2000 kg car is moving at 15 m/sec when it collides with a 1200 kg car sitting still. Briefly compare the impulse imparted on t

he 1200 kg car by the 2000 kg car to the impulse imparted on the 2000 kg car by the 1200 kg car. Which car undergoes the greater change in momentum? Briefly explain your response.
Physics
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

The 1200 kg car.

Explanation:

The 1200 kg car had bo momentum at the initial stage of the journey because it wasn't on motion. But after collision with the 2000 kg car it gained alot of momentum.

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Two positive point charges are 4.9cm apart. If the electric potential energy is 70.0 μJ, what is the magnitude of the force betw
ExtremeBDS [4]

Hi there!

Recall the following:

V \text{ (Electric Potential Energy) } = \frac{kq_1q_2}{r}\\\\F_E = \frac{kq_1q_2}{r^2}

k = Coulomb's Constant (Jm/C²)

q = Charge (C)
r = distance between charges (m)

To calculate the electric force between the two charges, we can simply divide by another 'r' (distance):

F_E = \frac{70}{0.049} = \boxed{1428.57 \mu J}

6 0
2 years ago
A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
PSYCHO15rus [73]

a) zinc

The equation of the photoelectric effect is:

E=\phi + K (1)

where

E is the energy of the incident light

\phi is the work function

K is the maximum kinetic energy of the emitted photoelectrons

Here the wavelength of the incident light is

\lambda=200 nm = 2\cdot 10^{-7} m

so the energy of the light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7} m}=9.95\cdot 10^{-19} J

Converting into electronvolts,

E=\frac{9.95\cdot 10^{-19}}{1.6\cdot 10^{-19} J/eV}=6.22 eV (2)

The stopping potential is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the electrical potential energy corresponding to the stopping potential (V=1.93 V) must be equal to the maximum kinetic energy of the photoelectrons,

U=q V = K

and since the charge of the electron is

1 q = 1 e

We have

K=(1 e)(1.93 V)=1.93 eV (3)

Combining (1), (2) and (3), we find the work function of the material:

\phi = E-K=6.22 eV-1.93 eV=4.29 eV

So, the cathode is most likely made of zinc, which has a work function of 4.3 eV.

b) The stopping potential does not change

As we said in part A), the stopping potential is proportional to the maximum kinetic energy of the photoelectrons, K.

The intensity of light is proportional to the number of photons that hit the surface of the metal. However, the energy of these photons does not depend on the intensity, but only on the frequency of the light.

Therefore, the energy of the photons (E) does not change when the intensity of light is doubled. Also, the work function \phi does not change: this means that the maximum kinetic energy of the photoelectrons, K, does not change, and so the stopping potential remains the same.

6 0
3 years ago
1 point
AysviL [449]
I think it’s 150 because if u multiply 3 x 50 it’s 150 but so it’s ore simple you do 3 x5 equals 15 then add the zero from the 50 so you’ll get 150 hope I helped
4 0
2 years ago
Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom
Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

B = B_o(-\hat k)

now the velocity is from bottom to top

so we have

v = v_o \hat j

now the force on the moving charge is given as

\vec F = q(\vec v \times \vec B)

now we have

\vec F = (-e)(v_o \hat j \times B_o(-\hat k))

\vec F = e v_o B \hat i

so force will be towards Right

7 0
3 years ago
Maria drove to the store and then to work.
natali 33 [55]

Answer: C - 60km !!!

Explanation:

8 0
2 years ago
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