Answer:
Explanation:
Given
mass of First Block
Temperature
mass of second block
Temperature
Heat capacity of aluminium c=899 J/kg-K
Final Temperature acquired by both blocks at steady state
Heat loss first block =Heat gain by second block
If the wavelength of the red line of hydrogen coming from a galaxy is 654.3 nanometers instead of the normal 656.3 nanometers, i
1 answer:
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<h2>The rank is : 
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We need to rank the objects according to the magnitude of their momentum.
We know momentum ,{ here m is mass and v is velocity }
Momentum of object A ,
Momentum of object B ,
Momentum of object C ,
Now, we can rank them in order of their magnitude of momentum .
.
Hence, this is the required solution.
Learn More :
Momentum
Answer:
Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s
Explanation:
Let's start out with finding the force acting downwards because of the mass of 'The Rock':
Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N
Now the problem also states that the kinetic friction of the desk in this problem is 370 N
Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N
Now lets use F = ma to calculate for the acceleration of the desk:
787.58 = 63 x acceleration
acceleration = 12.501 m/s
Finally, we can use the motion equation:
here u = 0 m/s (since initial speed of the desk is 0)
a = 12.501 m/s
and s = 10 m
Solving this we get:
Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at 15.812 m/s when the desk reaches the window.
Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U =
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ =
starting point.
Em₀ = K + U₁ + U₂
Em₀ =
final point
Em_f =
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( )
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ ) = 9 109 (1.6 10-19) ²(
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( )
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 =
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m