Answer:
All the three quantities will have non zero joules.
Explanation:
At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.
when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero
At the position of (y-h/2) all the 3 quantities will be non zero.
Answer:
D. provide the most compelling evidence of cause-and-effect relationships.
Explanation:
Answer:
A. The waves in the water travel faster and at a higher frequency than they travel on land.
Explanation:
The main reason why human ears can hear dolphins' vocalizations while under the water but cannot hear them well on land is because water is denser than air and air particles travel faster in denser particles.
Denser particles also ensures that the frequency of the waves move faster which in turn produces a faster and louder result.
Answer:
R = m⁴/kg . s
Explanation:
In this case, the best way to solve this is working with the units in the expression.
The units of velocity (V) are m/s
The units of density (d) are kg/m³
And R is a constant
If the expression is:
V = R * d
Replacing the units and solving for R we have
m/s = kg/m³ * R
m * m³ / s = kg * R
R = m * m³ / kg . s
<h2>
R = m⁴ / kg . s</h2>
This should be the units of R
Hope this helps
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
