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schepotkina [342]
3 years ago
11

HELP: When Jane drives to work, she always places her purse on the passenger's seat. By the time she gets to work, her purse has

fallen on the floor in front of the passenger seat. One day, she asks you to explain why this happens in terms of physics. What do you say?
Physics
1 answer:
Rudiy273 years ago
5 0

a force is a push or pull that acts upon an object as a result of its interaction with another object.  Newtons third law of motion suggests that something standing still will stay that way until another force pushes it.  This applies to Janes bag.  Every time Jane hits the brakes in her car, her bag will move forward until it finally falls off the seat.

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Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact w
aliya0001 [1]

Answer:

   f = 878,080 N

Explanation:

mass of pile driver (m) = 2100 kg

distance of pile driver to steel beam (s) = 5 m

depth of steel driven (d) = 12 cm = 0.12 m

acceleration due to gravity (g0 = 9.8 m/s^{2}

calculate the average force exerted on the pile driver by the beam.

  • from work done = force x distance
  • work done = change in potential energy of the pile driver
  • equating the two equations above we have

               force x distance = m x g x (s - d)

              f x 0.12 = 2100 x 9.8 x (5- (-0.12))

              d = - 0.12 because the steel beam went down at we are taking its  

              initial position to be an origin point which is 0

              f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12

                   f = 878,080 N

4 0
3 years ago
What is the average cost for a severe injury in a collision?
koban [17]

Answer:

D

Explanation:

According to research and data, the average cost for a severe injury in a collision is $247,000.

These include traumatic brain injury, severe damage to limbs which could result to loss of the limbs, spinal cord injuries that could result to partial or total paralysis as well as internal damage to organs in the collision.

8 0
3 years ago
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
3 years ago
50 POINTS!
zysi [14]

<h3>No:1</h3>

The object is moving with constant or uniform acceleration and in average speed

<h3>No:-2</h3>

The object is de accelerating

<h3>No:-3</h3>

The object deaccelerated and came to rest so fast.

<h3>No:-4</h3>

The object moves slowly first then accelerated.

<h3>No:-5</h3>

The object accelerated at first so fast then move with constant acceleration then again accelerated .

7 0
2 years ago
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