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valina [46]
3 years ago
13

At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de

vice to stop the train so that it will not plow through the station if the engineer misjudges the stopping distance. While waiting, you wonder what would be the fastest train that the spring could stop at its full compression which is L=3 ft . To keep the passengers safe when the train stops, you assume a maximum stopping acceleration of g/2. You also guess that a train weighs half a million lbs. For purpose of getting an estimate, you decide to assume that all frictional force are negligible.I've already posted this and someone gave me the wrong answer and i cannot for the life of me figure it out. The answer is not 7.66 m/s and should not be in meters, as the question gives everything else in feet.
Physics
1 answer:
xxMikexx [17]3 years ago
8 0

Answer:

  v₀ = 9,798 ft / s

Explanation:

We can solve this problem with kinematics in one dimension, when the train stops the speed is zero, the acceleration is negative so that the train stops. Let's use the equation

         v² = v₀² - 2 a d

         v = 0

         v₀ = √2 a d

In the problem it indicates that the acceleration is g / 2, we substitute

         v₀ = √2 (g / 2) d

Let's calculate

         v₀ = √ 2 32/2 3 = √32 3

         v₀ = 9,798 ft / s

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You and your friend are pushes hard against a stationary wall. If you push 3 times harder than your friend, then the amount of w
shtirl [24]

Answer:

Work = F * s    where s is the distance F moves

Since F is stationary, in this case, "no work" is done by either person

5 0
1 year ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
3 years ago
The fragment of an asteroid or any interplanetary material is known as a
Elodia [21]
The fragment of an asteroid or any interplanetary material is known as A. METEROID
8 0
3 years ago
Consider a system of two particles: ball A with a mass m is moving to the right a speed 2v and ball B with a mass 3m is moving t
arlik [135]

Answer:

Explanation:

Answer:

Explanation:

Given that,

System of two particle

Ball A has mass

Ma = m

Ball A is moving to the right (positive x axis) with velocity of

Va = 2v •i

Ball B has a mass

Mb = 3m

Ball B is moving to left (negative x axis) with a velocity of

Vb = -v •i

Velocity of centre of mass Vcm?

Velocity of centre of mass can be calculated using

Vcm = 1/M ΣMi•Vi

Where M is sum of mass

M = M1 + M2 + M3 +...

Therefore,

Vcm=[1/(Ma + Mb)] × (Ma•Va +Mb•Vb

Rearranging for better understanding

Vcm = (Ma•Va + Mb•Vb) / ( Ma + Mb)

Vcm = (m•2v + 3m•-v) / (m + 3m)

Vcm = (2mv — 3mv) / 4m

Vcm = —mv / 4m

Vcm = —v / 4

Vcm = —¼V •i

3 0
3 years ago
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