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Nadusha1986 [10]
4 years ago
15

When a clock reads 5.0 seconds a cart's velocity in the negative x direction is 3.0 m/s. when the clock reads 6.0 seconds, the c

art's velocity in the positive x direction is 4.0 m/s. what is the magnitude (in m/s2) of the average acceleration of the cart?
Physics
1 answer:
DIA [1.3K]4 years ago
7 0
Impulse is the change of momentum before and after the collision:
I = mv₁ - mv₂ = m (v₁-v₂)
m mass
v velocity

Impuls also is defined as the average force during a short period of time:
I = F * t = m * a * t
F average force
t time difference
m mass
a average acceleration

Combined:
m(v₁- v₂) = m * a * t
a = (v₁- v₂) / t
= (4m/s - (-3m/s)) / 6s - 5s
= 7m/s / 1s
= 7 m/s²


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Assuming the earth and the sun to be perfect spheres,

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8 0
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A pitcher throws an overhand fastball from an approximate height of 2.65 m and at an angle of 2.5° below horizontal. The catcher
rodikova [14]

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 = u_y·t + 1/2·g·t², we have;

u_y = 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 = u_y·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.

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