The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.
<h3>Final speed of the nickel</h3>
Apply the principle of conservation of energy.
Q = mcΔθ
Q = (18)(0.444)(66 - 20)
Q = 367.63 J
Q = K.E = ¹/₂mv²
2K.E = mv²
v = √(2K.E/m)
where;
v = √(2 x 367.63)/(0.018))
v = 202.1 m/s
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Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision
Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r = 
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r = 
= 0.0328 m, or 3.28 cm.
Answer:
The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.
See attachment below for full solution
Explanation:
This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.
<u>The two ways to find acceleration in non uniform motion are as follows:</u>
<u>Explanation:</u>
Non-uniform acceleration comprises the most common description of motion. Acceleration refers to the rate of changes of velocity per unit time. Basically, it implies that acceleration changes during motion. This variety can be communicated either as far as position (x) or time (t).
Accordingly, non-uniform acceleration motion can be carried out in 2 ways:
Calculus analysis is general and accurate, but limited to the availability of speed and acceleration expressions. It is not always possible to get the expression of motion attributes in the form "x" or "t". On the other hand, the graphic method is not accurate enough, but it can be used accurately if the graphic has the correct shapes.
The use of calculations involves differentiation and integration. Integration enables evaluation of the expression of acceleration of speed and expression of movement at a distance. Similarly, differentiation allows us to evaluate expression of speed position and expression speed to acceleration.
