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Goshia [24]
3 years ago
5

Martha must carry a 45 N package up three flights of stairs. Each flight of stairs has a height of 2.3m, and the actual distance

of a diagonal path she walks up the stairs is 10 m. What is the total work done on the package?
Physics
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

310.5 J

Explanation:

The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to

\Delta U = mg\Delta h

where

(mg) = 45 N is the weight of the package

\Delta h is the increase in height of the package

The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is

\Delta h = 3 \cdot 2.3 m=6.9 m

And so, the work done by Martha is

U=(45 N)(6.9 m)=310.5 J

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All objects, regardless of their mass, fall with the same rate of acceleration on
tamaranim1 [39]

Answer:

A -TRUE

Explanation:

The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration.

4 0
2 years ago
Read 2 more answers
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
Who has the greater velocity, an astronaut who has just completed an orbit of the Earth or you when you have just traveled from
Alik [6]

Answer:

the answer is

Explanation:

constant acceleration

because when the object's velocity is changing then the object is accelerating or decelerating

as acceleration describe changing of velocity so the answer is constant acceleration

Acceleration is defined as the rate of change of velocity.

Acceleration = (Change in velocity) / time taken

Acceleration = (Final velocity - initial velocity) / time

As the object velocity changes by the same amount in each second, it means the acceleration is constant.

Hope I can help u

4 0
2 years ago
How much does the speed of a car increase if it accelerates
Nata [24]

Answer:

12.5 m/s

Explanation:

a = Δv / Δt

2.5 m/s² = Δv / 5 s

Δv = 12.5 m/s

4 0
3 years ago
What is the slope of the line plotted below?<br> A. 1.33<br> B. -1.33<br> C. -0.6<br> D. 0.6
Naya [18.7K]

Answer:

C. -0.6

Explanation:

Line is passing through the points ( - 3, 1) & (2, - 2)

Slope of line

=  \frac{ - 2 - 1}{2 - ( - 3)}  =  \frac{ - 3}{2 + 3}  =  \frac{ - 3}{5}  =  - 0.6 \\

4 0
3 years ago
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