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3 years ago

“A horse pulls on a cart. By Newton’s third law of motion, the cart pulls back on

2 answers:

Studentka2010 [4]3 years ago

8 0

In order to see what's going on, let's put them in empty space to get rid of any other influences, and let's also make it a push instead of a pull. / / / The horse pushes on the cart, so it begins accelerating away from him. At the same time, because of the equal opposite reaction thing, the cart pushes back on the horse, so the horse starts accelerating backwards, away from the cart. They both accelerate in opposite directions from where they started. BUT . . . their common center of mass doesn't move, and the sum of their momentums (which are in opposite directions) remains zero.

Slav-nsk [51]3 years ago

4 0

If they are both in space and not on ground, then the above way of thinking is correct. Horse will not be able to accelerate forward.

But here, there are some external forces acting on the horse other than the pull of the cart. That makes the horse accelerate. Cart also gets accelerated.

Horse kicks and pushes the road or ground with his feet backwards. So the ground pushes with an equal force in the forward direction. On the horse, there is a force by the cart, pulling it backwards and a force of friction from the ground and a push forward given by the ground. Pushing forward force by the ground is higher than the sum of others. So horse accelerates.

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A chemical reaction takes place in which energy is absorbed. Arrange the characteristics of the reaction in order from start to

Ira Lisetskai [31]

<h2>Answer</h2>

It will be single step endothermic reaction.

<h2>Expalantion</h2>

In the endothermic reaction, the reactants come together to convert to products by absorbing the heat from the external source. This reaction is explained is also known as one step reaction as reactants meet to get the transition stage and converts to the product. But in some reactions, the activation energy required to activate the reactants to get the transition stage to form products. For this, the reaction will have the steps as activation energy, reactant meet, transition stage and products form.

6 0

3 years ago

Read 2 more answers

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0

3 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

3 years ago

Answer it or not your choice

Zigmanuir [339]

Answer:

Explanation:

(a) 0 to 90

(b) 90 to 180

3 0

2 years ago

A train takes 2h to reach station B from station A and 3h to return back to A.The distance between the station is 200km, then it

serg [7]

Answer:

80km/hr

Explanation:

Total distance: 400km

Total time 5 hours

Average speed= distance/time

400/5= 80km/hr

8 0

3 years ago