Answer:
A) m2 = 98.71g
B) v_f2 = 1.86 m/s
Explanation:
We are given;
Mass of cart; m1 = 340g
Initial speed; v_i1 = 1.2 m/s
Final speed; v_f1 = 0.66 m/s
A)Since the collision is elastic, we can simply apply the conservation of momentum to get;
m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)
From conservation of kinetic energy, we have;
(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)
Let's make v_f2 the subject in eq 2;
Thus,
v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)
v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)
Let's put this for v_f2 in eq1 to obtain;
m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)
Let's square both sides to give;
(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]
This gives;
m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]
Plugging in the relevant values to get;
m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]
m2 = 98.71g
B) from equation 1, we have;
m1•(v_i1) = m1•(v_f1) + m2•(v_f2)
Making v_f2 the subject, we have;
v_f2 = m1[(v_i1) - (v_f1)]/m2
Plugging in the relevant values to get;
v_f2 = 340[(1.2) - (0.66)]/98.71
v_f2 = 1.86 m/s
To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final velocity of the combined carts to the initial velocity of the first cart.
To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.
After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.
#SPJ11
The measurement will be significantly affected.
Recall that the relationship between linear velocity and angular velocity is subject to the formula
,
Where r indicates the radius and the angular velocity.
As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.
True or false
Answer:
False
whenever the string breaks, the ball will follow the straight line tangential path
Explanation:
No, the ball will not follow a curved path after the string breaks. Since, the the direction of velocity is tangential to each point of the circular motion. Therefore, it changes at every point. This produces an acceleration in the circle called centripetal acceleration. There is also a tangential component of acceleration acting on the ball during this motion.
So, whenever the string breaks, the ball will follow the straight line tangential path. Hence, the given statement is false.
Answer:
U/U₀ = 2
(factor of 2 i.e U = 2U₀)
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected
Explanation:
Energy stored in a capacitor can be expressed as;
U = 0.5CV^2 = Q^2/2C
And
C = ε₀ A/d
Where
C = capacitance
V = potential difference
Q = charge
A = Area of plates
d = distance between plates
So
U = Q^2/2C = dQ^2/2ε₀ A
The initial energy of the capacitor at d = d₀ is
U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1
When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.
The final energy stored in the capacitor at d = 2d₀ is
U = 2d₀Q^2/2ε₀ A ...2
The factor U/U₀ can be derived by substituting equation 1 and 2
U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )
Simplifying we have;
U/U₀ = 2
U = 2U₀
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
Answer:
The answer is below
Explanation:
The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:
V² = 56² + 126²
V² = 19012
V = 137.9 m/s
Answer:
For 25-turn electromagnet, Number of clips = 4.1
For 50-turn electromagnet number of clips = 9.6
Explanation:
To calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.
Hence;
Using the equations gotten from the graph in the previous question and 1.0 V as the value for x, we get
For 25-turn electromagnet y = 3.663x * 0.5
(rounded to one decimal place) Number of clips = 4.1
For 50-turn electromagnet y = 7.133x 2.5
(rounded to one decimal place) Number of clips = 9.6