What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure
1 answer:
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Answer:
The initial velocity of the pitch is approximately 36.5 m/s
Explanation:
The given parameters of the thrown fastball are;
The height at which the pitcher throws the fastball, h₁ = 2.65 m
The angle direction in which the ball is thrown, θ = 2.5° below the horizontal
The height above the ground the catcher catches the ball, h₂ = 1.02 m
The distance between the pitcher's mound and the home plate = 18.5 m
Let 'u' represent the initial velocity of the pitch
From h = ·t + 1/2·g·t², we have;
= The vertical velocity = u·sin(θ) = u·sin(2.5°)
h = 2.65 m - 1.02 m = 1.63 m
uₓ·t = u·cos(θ) = u·cos(2.5°) × t = 18.5 m
∴ t = 18.5 m/(u·cos(2.5°))
∴ h = ·t + 1/2·g·t² = (u·sin(2.5°))×(18.5/(u·cos(2.5°))) + 1/2·g·t²
1.63 = 8.5·tan(2.5°) + 1/2 × 9.8 × t²
t² = (1.63 - 8.5·tan(2.5°))/(1/2 × 9.8) = 0.25691469087
t = √(0.25691469087) ≈ 0.50686752763
t ≈ 0.50686752763 seconds
u = 18.5 m/(t·cos(2.5°)) = 18.5 m/(0.50686752763 s × cos(2.5°)) = 36.5334603 m/s ≈ 36.5 m/s
The initial velocity of the pitch = u ≈ 36.5 m/s.
Answer:
We conclude that the required momentum is 80 kgm/s.
Explanation:
Given
- Mass m = 20 kg
- Velocity v = 4 m/s
To determine
The momentum p = ?
Important Tip:
- The momentum of an object is the product of mass and velocity.
We can determine the momentum using the formula
where
- p is the momentum
- m is the mass
- v is the velocity
now substituting m = 20 and v = 4 using the equation
kg m/s
Therefore, we conclude that the required momentum is 80 kgm/s.
Answer:
The acceleration of the wallet is
Explanation:
Given that,
Radius of purse r= 2.30 m
Radius of wallet r'= 3.45 m
Acceleration of the purse
We need to calculate the acceleration of the wallet
Using formula of acceleration
Both the purse and wallet have same angular velocity
Hence, The acceleration of the wallet is