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Kruka [31]
3 years ago
7

A driver in a car traveling at a speed of 21.8 m/s sees a cat 101m away on the road. How long will it take for the car to accele

rate uniformly for 3.2 km in 3.5 min. How fast (in m/s) is the car moving after this time?
Physics
1 answer:
Ket [755]3 years ago
8 0

The answer would be 9.1 s


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Someone please help me
Schach [20]

Answer:

300x480 teaspoons

Explanation:

when converting cups to teaspoons just multiply by 48

3 0
3 years ago
Read 2 more answers
What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
serg [7]

Answer: 6.268(10)^{-16}J

Explanation:

The kinetic energy of an electron K_{e} is given by the following equation:

K_{e}=\frac{(p_{e})^{2} }{2m_{e}}   (1)

Where:

K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

m_{e}=9.11(10)^{-31}kg  is the mass of the electron

From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

\lambda_{e}=\frac{h}{p_{e}}    (4)

Where:

h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

So, we will use the value of p_{e} found in (3) for equation (4):

\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}    

\lambda_{e}=3.168(10)^{-10}m    (5)

We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}  

Finally:

E_{p}=6.268(10)^{-16}J    

4 0
3 years ago
The light reactions could be viewed as analogous to a hydro-electric dam. In that case, the wall of the dam that holds back the
Viktor [21]

The light reactions could be viewed as analogous to a hydroelectric dam. In that case, the wall of the dam that holds back the water would be analogous to the thylakoid membrane.

Thylakoid membrane is present in cyanobacteria and chloroplasts of plants. It plays a crucial role in photosynthesis and photosystem II reactions.

In general, these are the regions where light-dependent reactions take place. The thylakoid membrane is a lipid-bound membrane that maintains potential difference and also controls the flow of liquids across the membrane during light reactions.

In the provided case, we can see that the wall of the dam holds back the water, similarly, in light-dependent reactions, thylakoid membranes control the liquid flow and also regulate the potential gradient across the membrane and also allow the selective proteins to pass through.

If you need to learn more about light reactions click here:

brainly.com/question/26623807

#SPJ4

3 0
1 year ago
We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength
san4es73 [151]

Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

\theta=1.22\frac{\lambda}{b}

θ = 50.0*10^-7 rad

λ: wavelength of the light = 550nm

b = diameter of the objective

By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm

hence, the smallest diameter objective lens is 13.4cm

8 0
3 years ago
Read 2 more answers
What is the unit of G in the F=Gm1m2/r^2​
kobusy [5.1K]


G
has the SI units
m
3
k
g
⋅
s
2
6 0
3 years ago
Read 2 more answers
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