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Nataly [62]
3 years ago
8

Which skateboarder has greater momentum?

Physics
1 answer:
Verizon [17]3 years ago
5 0

Answer:

skateboard b

Explanation:

p=mv

skateboard a

p=(60kg)(1.5m/s)=90kg*m/s

skateboard b

p=(50kg)(2m/s)=100kg*m/s

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emmainna [20.7K]
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3 0
3 years ago
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We can see Objects because of
Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

7 0
3 years ago
Compare and contrast instantaneous and average speed.
Harlamova29_29 [7]

Answer:

instantaneous velocity is a velocity covered at an instant while average velocity is the change in distance/ the change in time taken

6 0
2 years ago
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?
Komok [63]
Data:

E = 20 V
R_{1} = 65\Omega
R_{2} = 65\Omega
R_{3} = 65\Omega
R_{4} = 65\Omega

<span>Now that we have all the values ​​we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

</span>R_{T} =  R_{1} + R_{2} + R_{3} + R_{4}
R_{T} = 65 + 65 + 65+ 65
R_{T} = 260\Omega

<span>Like this:
</span>
I_{T} =  \frac{E}{ R_{T} }

I_{T} = \frac{20}{ 260 }
I_{T} = 0,076923076...

\boxed{\boxed{I_{T} \approx 0,07A}} 
Answer:
<span>The intensity of the total electric current 
</span>\boxed{\boxed{I_{T} \approx 0,07A}} 

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.
4 0
3 years ago
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