First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency
Cardio respiratory endurance, muscular strength, muscular endurance
The thin atmosphere of Mars is thought to be due to the planet's lack of a magnetic field, which has allowed the Solar wind to blow away much of the gas the planet once had. Venus, despite still having a thick atmosphere of CO2, surprisingly has a similar problem
Answer:
![\frac{KE_{Rotational}}{KE_{Total}} = 0.018](https://tex.z-dn.net/?f=%5Cfrac%7BKE_%7BRotational%7D%7D%7BKE_%7BTotal%7D%7D%20%3D%200.018)
Explanation:
To develop this exercise we proceed to use the kinetic energy equations,
In the end we replace
![KE_{Total}=KE_{Translational}+KE_{Rotational}](https://tex.z-dn.net/?f=KE_%7BTotal%7D%3DKE_%7BTranslational%7D%2BKE_%7BRotational%7D)
![KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2](https://tex.z-dn.net/?f=KE_%7BTotal%7D%3D%5Cfrac%7B1%7D%7B2%7Dm_%7Bcar%7D%2B4%2A%5Cfrac%7B1%7D%7B2%7D%2AI%2A%28%5Cfrac%7Bv%7D%7Br%7D%29%5E2)
Here
meaning the 4 wheels,
So replacing
![KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2](https://tex.z-dn.net/?f=KE_%7BRotational%7D%3D4%5Cfrac%7B1%7D%7B2%7D%2A%28%5Cfrac%7B1%7D%7B2%7Dm_%7Bwheels%7D%2Ar%5E2%29%2A%28%5Cfrac%7Bv%7D%7Br%7D%29%5E2%3Dm%2Av%5E2)
So,
![\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}](https://tex.z-dn.net/?f=%5Cfrac%7BKE_%7BRotational%7D%7D%7BKE_%7BTotal%7D%7D%20%3D%20%5Cfrac%7Bm_%7Bwheels%7D%2Av%5E2%7D%7B%5Cfrac%7B1%7D%7B2%7Dm_%7Bcar%7D%2Av%5E2%2Bm_%7Bwheels%7D%2Av%5E2%7D)
![\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}](https://tex.z-dn.net/?f=%5Cfrac%7BKE_%7BRotational%7D%7D%7BKE_%7BTotal%7D%7D%20%3D%20%5Cfrac%7Bm_%7Bwheels%7D%7D%7B%5Cfrac%7B1%7D%7B2%7Dm_%7Bcar%7D%2Bm_%7Bwheels%7D%7D)
![\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}](https://tex.z-dn.net/?f=%5Cfrac%7BKE_%7BRotational%7D%7D%7BKE_%7BTotal%7D%7D%20%3D%20%20%5Cfrac%7B10%7D%7B545%2B10%7D)
![\frac{KE_{Rotational}}{KE_{Total}} = 0.018](https://tex.z-dn.net/?f=%5Cfrac%7BKE_%7BRotational%7D%7D%7BKE_%7BTotal%7D%7D%20%3D%200.018)
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