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sashaice [31]
3 years ago
14

A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest.

What is the common velocty of the two-car train after the collisions?
Physics
1 answer:
Tomtit [17]3 years ago
6 0

Answer:

v = 6.79 m/s

Explanation:

It is given that,

Mass of a train car, m₁ = 11000 kg

Speed of train car, u₁ = 21 m/s

Mass of other train car, m₂ = 23000 kg

Initially, the other train car is at rest, u₂ = 0

It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,

m_1u_1+m_2u_2=(m_1+m_2)V

V is the common velocity after the collisions

V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s

So, the two car train will move with a common velocity of 6.79 m/s.

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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq
gayaneshka [121]

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law,  PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}

Given;

\frac{dP}{dt} (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) =  600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( \frac{dv}{dt});

(600 x 40) + (150 x \frac{dv}{dt}) = 0

\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min

Therefore, the volume is decreasing at 160 cm³/min

3 0
3 years ago
"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat
vagabundo [1.1K]

Answer:

(a) g = 8.82158145m/s^2.

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

g = \frac{M_{(earth)} }{r^2} G , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145m/s^2.

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

a_{centripetal}=\frac{V^2}{r} =g here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c)  S = vT,  here T is time period or time required to complete one full revolution.

S =  earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

3 0
3 years ago
A marble is dropped from rest and falls for 2.3 seconds. Find its final velocity.
juin [17]

Answer:

23 m/s downward

__________________________________________________________

<em>Taking the downward direction as positive</em>

<u>We are given:</u>

Initial velocity of the marble (u) = 0 m/s

Time interval (t) = 2.3 seconds

Final velocity (v) = x m/s

<u>Solving for the Final velocity:</u>

<u>Acceleration of the Marble:</u>

We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s

<u>Final velocity:</u>

v = u + at                                              [First equation of motion]

x = 0 + (10)(2.3)                                    [replacing the given values]

x = 23 m/s

Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction

4 0
3 years ago
In the fusion reaction occurring in the Sun, how does the mass of helium atom formed compare to the mass of the hydrogen atoms?
PolarNik [594]
B bc there is a lot of hydrogen used

3 0
3 years ago
If the Earth rotated in the opposite direction, how would that affect the wind? *
Alik [6]

Answer:

The wind would still blow, but it would curve and spin in the opposite direction.

Explanation:

6 0
3 years ago
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