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Gwar [14]
2 years ago
6

A bicyclist is traveling 10 m/s at 30 degrees

Physics
1 answer:
AysviL [449]2 years ago
8 0

Explanation:

The momentum is defined as the product between the mass and velocity:

p = m·v

Let's calculate his total momentum:

ptot = 70kg · 10m/s = 700 kg·m/s

Now we know that he's traveling 30 degrees south-east.

The momentum along the y axis will be:

py = p·cos30° = 606 kg·m/s

Along the x-axis:

px = psin30° = 350 kg·m/s

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An airplane is being pushed West by its engines with a force of 350 N as it starts to take off the runway at the Atlanta airport
Marizza181 [45]

Answer:

option C

Explanation:

given,

Force by the engine on plane in West direction = 350 N

Frictional force on the runway = 100 N in east

force exerted by the wind = 100 N in east

net force and direction = ?

consider west to be positive and east be negative.

when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane

Net force = 350 N - 100 N - 100 N

                 = 150 N

as our answer comes out to be positive so the airplane will be moving in West

hence, the correct answer is option C

8 0
2 years ago
What best describes an impulse acting on an object? the net force on an object divided by the time of impact the velocity of an
Leona [35]

Answer:

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Explanation:

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7 0
2 years ago
Balanced forces acting on an object cause the object to accelerate. Please select the best answer from the choices provided T F
Pani-rosa [81]
False
Balanced forces mean that there is no net force acting on the object. therefore, the object will not accelerate.
7 0
3 years ago
The oceans level is currently rising at about 2 mm per year at this rate in how many years will sea level be 3 m higher than now
lesantik [10]
(3 m) / (2 mm/yr) = (3,000mm)/(2mm/yr) = 1,500 yrs.
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5 0
3 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
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