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LiRa [457]
3 years ago
13

A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid

er together is 85 kg, what is the net force acting on the bicycle? (Hint: Calculate Acceleration!)
Physics
1 answer:
Gnom [1K]3 years ago
4 0

Answer:

42.5 N

Explanation:

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What will happen if you heat a liquid to high temperatures?
WARRIOR [948]
It’s c hope this helps :)
5 0
3 years ago
TECHNOLOGY A Lamborghini Aventador accelerates from rest to 100 km / h in 2.9 s. The mass of the car is 1575 kg Calculate the po
DerKrebs [107]

At 100 km/hr, the car's kinetic energy is

KE = (1/2) (mass) (speed)²

KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²

KE = (787.5 kg) (27.78 m/s)²

KE = 607,639 Joules

In order to deliver this energy in 2.9 seconds, the engine must supply

(607,639 J / 2.9 sec) = 209,531 watts

<em>Power = 281 HP</em>

6 0
3 years ago
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
4 years ago
Please help
vladimir2022 [97]

Answer:

true true false true false

5 0
2 years ago
Read 2 more answers
what is the position of centre of curvature for concave and convex mirror show with the help of diagram if you can​
Anon25 [30]

it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram

8 0
3 years ago
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