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Wewaii [24]
2 years ago
8

An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar

has a mass of 38500 kg and is moving to the right at a constant speed of 8.70 m/s on a frictionless rail. The automobile them accelerates to the left, leaving the railcar at a speed of 22.0 m/s with respect to the ground. When the automobile lands, what is the distance D between it and the left end of the railcar?
Physics
1 answer:
igomit [66]2 years ago
7 0
<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
</span>
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A car’s momentum is p when it is traveling with a velocity of v. If the velocity of that car doubles, what is the new momentum o
lubasha [3.4K]

Answer:

∆p=(m2v)kg.m/s

Explanation:

∆p=mv where v=2v. hence ∆p=m2v

5 0
2 years ago
Read 2 more answers
Consider an unknown charge that is released from rest at a particular location in an electric field so that it has some initial
sineoko [7]

Answer:

b)

Explanation:

If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.

For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)

For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)

4 0
3 years ago
Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
kirill [66]

Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²\sqrt{((17.4)^{2} +(16.9)^{2}} = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

4 0
3 years ago
Which nuclei is not radioactive? <br> A. Am-241<br> B. Mg-24<br> C. Pu-241<br> D. U-238
Helen [10]

Hello!

Which nuclei is NOT radioactive?

A) Am-241 B) Mg-24 C) Pu-241 D) U-238

Solving:

It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.  

Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.  

Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.

Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.  

U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.

Answer:

B) Mg-24

_______________________

I Hope this helps, greetings ... Dexteright02! =)

6 0
3 years ago
What is the magnitude of your displacement when you follow directions that tell you to walk 100.0m north, then 25.0m East?
atroni [7]

Answer:

Displacement from the starting position is 103.21m

Explanation:

If you draw these directions, it will create the two legs of a triangle.

Using this method, you can visualize why your displacement is what it is.

Using the pythagorean theorem

{a}^{2}  +  {b}^{2}  =  {c}^{2}

Plug in both values

{100m}^{2}  +  {25m}^{2}  =  {c}^{2}

c =  \sqrt{10652}

c = 103.2085

c= 103.21

6 0
3 years ago
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