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Mazyrski [523]
3 years ago
15

A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same

direction, and the two latch onto each other and continue together. What is their final speed?
A.) 16 m/s
B.) 6.0 m/s
C.) 10 m/s
D.) 14 m/s
Physics
1 answer:
emmainna [20.7K]3 years ago
6 0
This item is solved through the concept of the conservation of momentum which states that the momentum before and after collision should be equal. 
                                   momentum = mass x velocity
            (1,600 kg)(16 m/s) + (1.0x10^3 kg)(10 m/s) = (1600 + 1000 kg)(x)
The value of x is 13.69 m/s. Thus, their final speed is approximately letter D. 14 m/s. 
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A fox with a thick fur would have a survival advantage over other foxes if
Nat2105 [25]
This question is poorly stated, but I assume you mean what conditions are needed. It would have to be cold outside, correct?
8 0
3 years ago
In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of
antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, a=84\ au= 1.496\times 10^{11}\ m

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e T^2=\dfrac{4\pi^2}{GM}a^3

where G is gravitational constant

M is  mass of sun, M=1.98\times 10^{30}\ Kg

So, T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}

T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3

T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}

T^2=59532469.8\times 10^{14}\ s

T=7715.7\times 10^7\ s

since, 1\ sec=3.17\times 10^{-8}\ years

So, orbital period is approximately 2446 years.

7 0
3 years ago
If one object is 103 km away and a second object is 106 km away, one could say that the second object is _____ times further awa
vaieri [72.5K]

Answer:

1.03

Explanation:

\frac{object_{second}}{object_{first}} = \frac{106}{103} = 1.02912621359

Round to three significant digits

1.03

7 0
3 years ago
What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?
Taya2010 [7]

Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:

Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Answer:

The Force act on each side is 2.43×10⁴N

Explanation:

Given data

n=3 mol

L=0.3 m

Temperature=20.0°C=293 K

To find

Force F

Solution

To get force act on each side it would employ by

F=P.A

Where P is pressure

A is Area

First we need to find pressure by applying ideal gas law

So

P.V=nRT\\P=\frac{nRT}{V}\\ P=\frac{(3mol)(8.315J/mol)(293K)}{(0.3m*0.3m*0.3m)}\\P=27.069*10^{4}Pa

So The Force is given as:

F=P.A\\F=(27.069*10^{4} )(0.3m*0.3m)\\F=2.43*10^{4}N

The Force act on each side is 2.43×10⁴N

3 0
3 years ago
Para cargar un camión, se suele utilizar una tabla entre el contenedor y el suelo, con el fin de subir la carga, ya sea desplaza
stiks02 [169]

Answer:

A. No

B. si

Explanation:

A. El trabajo realizado en la carga es la energía potencial ganada por la carga al elevar la carga al nivel del camión y colocar la carga dentro del camión.

El trabajo realizado para elevar la carga W = m × g × h

Dónde;

m = masa de la carga

g = aceleración debido a la gravedad

h = Nivel de altura donde se coloca la carga en el camión

Por lo tanto, el trabajo realizado depende de la masa, m, de la carga y el nivel de altura, h, donde la carga se coloca en el camión y el trabajo realizado es el mismo para todos los métodos utilizados para colocar la carga en el camión

B. La ecuación para el trabajo realizado, W, también se puede escribir de la siguiente manera;

W = Fuerza, F × Distancia, D

De lo que tenemos;

F = W/D

Por lo tanto, cuando la mesa aumenta la distancia, como una rampa o un plano inclinado, la fuerza requerida disminuirá.

4 0
3 years ago
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