Given:
A cylindrical container closed of both end has a radius of 7cm and height of 6cm.
Explanation:
A.) Find the total surface area of the container.
- A = 2πrh + 2πr²
- A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
- A = 263.76 + 307.72
- A = 571.48
B.) Find the volume of the container.
- V = πr²h
- V = (3.14)(7×7)(6)
- V = 923.16
Not sure huhuness.
#CarryOnLearning
Answer:
The answer is 3.48 seconds
Explanation:
The kinematic equation
y= y0+V0*t+1/2*a*(t*t)
-50=0+(0)t+1/2(-9.8)*(t*t)
t=3.194 seconds
During ribbons ball,
x=x0+ Vt+1/2*a*(t*t)
x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)
x= 47.9157m
So, distance (D) = 100-47.9157= 52.084m
52.084m=0+15(t)+1/2*(0)(t*t)
t=52.084/15=3.472286= 3.48seconds
<span>The primary gases of the atmosphere are _____:
</span>
nitrogen, oxygen, and carbondioxide
We will apply the conservation of linear momentum to answer this question.
Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Note that v₁ and v₁' is the velocity of m₁ before and after the collision.
Let's choose m₁ and v₁ to represent the bullet's mass and velocity.
m₂ and v₂ represents the wood block's mass and velocity.
The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'
m₁v₁ + m₂v₂ = m₁v' + m₂v'
m₁v₁ + m₂v₂ = (m₁+m₂)v'
Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.
m₁v₁ = (m₁+m₂)v'
Here are the given values:
m₁ = 0.005kg
v₁ = 500m/s
m₂ = 5kg
Plug in the values and solve for v'
0.005×500 = (0.005+5)v'
v' = 0.4995m/s
v' ≅ 0.5m/s
Answer:
Because there is nothing out in space , the sound waves from one astronaut's whistling can't travel over to the other astronaut's ears.
