Answer:
my opi
Explanation:
Mass is like the inside of the weight. weight is just home much it weighs
Answer:
0.5A
Explanation:
Using 
,
R is the resistance (in Ohms)
V is the voltage (in V)
I is the current (in A)
I = 0.5A
Answer:
Explanation:
The form of Newton's 2nd Law that we use for this is:
F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).
We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:
That's everything we need.
w is weight: 6.0(9.8). Filling in:
6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and
2.0 × 10¹ - 8.8 = 6.0a and
11 = 6.0a so
a = 1.8 m/s/s
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
