Maurice directs a beam of light on a sheet of glass at an angle of 51°. The refractive index of glass is <span>1.46. </span>The angle of refraction in the glass is 29 degrees. The answer is letter B.
Answer:
K' = 1777.777 J
Explanation:
Given that
m = 40 kg
v= 15 m/s
K=1000
Given that kinetic energy(K) varies with mass(m) and velocity(v)
K= C(mv²)
Where
C= Constant
m=mass
v=velocity
When
m = 40 kg ,v= 15 m/s ,K=1000
K= C(mv²)
1000 = C( 40 x 15²)
C=0.111111
When m = 40 kg and v= 20 m/s
K' = C(mv²)
K= 0.1111 x (40 x 20²)
K' = 1777.777 J
Answer: Things continue doing what they are doing unless a force is applied to it. Objects have a natural tendency to resist change. This is INERTIA. Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects, so true
Explanation:
Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop. The Cadillac has more of a tendency to stay stationary (or continue moving), and resist a change in motion than a bicycle.
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
The statements that are held true with regards to the static equilibrium of bodies are:
<span>The net torque acting on the object must equal zero
</span><span>The net torque on the object does not have to be zero if the net force on the object is zero
Furthermore, when a body is in a state of static equilibrium, the summation of all forces, either vertically or horizontally, must be equal to zero. </span>
