Answer:the final speed is 5.01 m/s
Explanation:
Momentum is the product of mass and velocity.
Cart 1 has a mass of 4.2kg and a speed 5.4 m/s
Cart 2 has a mass of 3.2kg and a speed 4.5 m/s
Total momentum before collision is
m1u1 + m2u2. It becomes
4.2×5.4 + 3.2×4.5 = 22.68 + 14.4
= 37.08kgm/s
The carts stick together after colliding head-on. This means that they move with a common velocity, v. Therefore, Total momentum after collision is (m1 + m2)v. It becomes
(4.2 + 3.2)v = 7.4v
According the the law of conservation of momentum, the total momentum before collision = the total momentum after collision. Therefore,
7.4v = 37.08
v = 37.08/7.4 = 5.01 m/s
Answer:
the average induced emf in the coil is 0.016 V.
Explanation:
Given;
diameter of the wire, d = 14.1 cm = 0.141 m
change in magnetic filed strength, dB = 0.52 T - 0.23 T = 0.29 T
change in time, dt = 0.28 s
The area of the wire is calculated as follows;

The induced emf is calculated as follows;

Therefore, the average induced emf in the coil is 0.016 V.
Answer:
The gain in velocity is 0.37m/s
Explanation:
We need solve this problem though the conservation of momentum. That is,


Using the equation to find
,

Using the conservation of energy equation, we have,




Now this energy over the cannonball



The gain in velocity is 0.37m/s
Hi there! :)

Use the following equation in solving for kinetic energy:
KE = 1/2mv² where:
KE = kinetic energy (J)
m = mass (kg)
v = velocity (m/s)
Plug in the given values:
12,000 = 40v²
Divide both sides by 40:
12,000 / 40 = v²
300 = v²
Take the square root of both sides:
√300 = v
v ≈ 17.32 m/s
Answer:
<em>4.79m/s</em>
Explanation:
According to law of conservation of momentum;
The sum of momentum of the bodies before collision is equal to the momentum after collision.
m1u1 + m2u2 = (m1+m2)v
Given;
m1 = 0.3kg
u1 = 2.30m/s
m2 = 0.0225kg
u2 = 38m/s
Required
speed of the arrow after passing through the target v
Substituting the given data into the formula
0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v
0.69 + 0.855 = 0.3225v
1.545 = 0.3225v
v = 1.545/0.3225
v = 4.79m/s
<em>Hence the speed of the arrow after passing through the target is 4.79m/s</em>