Answer:
D
Explanation:
I would say much too small as there is a significant portion of the plant that is UNDER water.....
( Really depends on how deep the water is...if it is shallow water it would be just a little too small)
Answer: the mass of the second ball is 2.631 kg
Explanation:
Given that;
m1 = 0.877 kg
Initial velocity = V0
Initial momentum = m1 × V0
final velocity of m1 is u1, final velocity of m2 is u2 = v0/2
now final momentum = m1 × u1 + m2 × u2
using momentum conservation;
m1×V0 = m1×u1 + m2×v0/2
m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1
Now, for elastic collision;
m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2
m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2
now; equation 2 / equation 1
: V0 + u1 = v0/2
2V0 + 2u1 = V0
2u1 = V0 - 2V0
u1 = -V0/2
now we insert in equ 1
m1×3V0/2= m2×V0/2
m1 × 3 = m2
m2 = 0.877 × 3
m2 = 2.631 kg
Therefore, the mass of the second ball is 2.631 kg
Answer:
0.014 kg
Explanation:
trust me its right i took the assessment and got 100%. :)
Answer:
<h2>31km/h</h2>
Explanation:
Step one:
given data
speed of bus= 35km/h forward
speed of ball= 4km/h backwards
`Step two:
Required
the magnitude of the velocity of the ball relative to the ground is the net velocity
that is
=35-4
=31km/h
<em><u>The velocity of the ball relative to the ground is 31km/h</u></em>