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Mumz [18]
2 years ago
7

a boy throws a ball straight up into the air. it reaches its highest point after 4 seconds.how fast was the ball going when it l

eft the boy's hand​
Physics
1 answer:
baherus [9]2 years ago
3 0

Answer:

Explanation:

The most important thing to remember about parabolic motion in physics is that when an object reaches its max height, the velocity right there at the highest point is 0. Use this one-dimensional motion equation to solve this problem:

v = v₀ + at and filling in:

0 = v₀ + (-9.8)(4.0) **I put in 4.0 for time so we have more than just 1 sig fig here**

0 = v₀ - 39 and

-v₀ = -39 so

v₀ = 39 m/s

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Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert
chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s

Converting to km/h

100.66924\times 3.6=362.41\ km/h

The terminal velocity of the stone is 362.41 km/h

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3 years ago
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How many hydrogen and carbon atoms in a diamond
Wewaii [24]

Answer:

Explanation:

Thus, total 4+4=8 C atoms are present per unit cell of diamond. Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.

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3 years ago
Provide an example of when momentum is conserved and explain your answer you can get 10 PTS if answered with a good explaination
dezoksy [38]

Answer:

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

Explanation:

<u>Conservation of Momentum </u>

The total momentum of a system of two particles is

p=m_1v_1+m_2v_2

Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

p'=m_1v_1'+m_2v_2'

The momentum is conserved if no external forces are acting on the system, thus

m_1v_1+m_2v_2=m_1v_1'+m_2v_2'

Let's put some numbers in the problem and say

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

(8)(12)+(6)(4)=(8)(-6)+(6)(28)

96+24=-48+168

120=120

It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s

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