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3 years ago

A car traveling at 23m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is the acceleration?

1 answer:

7 0

The car has an initial velocity $v_0$ of 23 m/s and a final velocity $v$ of 0 m/s. Recall that for constant acceleration,

$v=v_0+at$

The car stops in 7 s, so the acceleration is

$0\,\dfrac{\mathrm m}{\mathrm s}=23\,\dfrac{\mathrm m}{\mathrm s}+a(7\,\mathrm s)$

$\implies a\approx-3.29\,\dfrac{\mathrm m}{\mathrm s^2}$

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3 years ago

A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 95 cm and x2 = 135 cm. (i

uranmaximum [27]

Answer:

(i) $x = 115\,cm$ , (ii) $x = 95\,cm$ , (iii) $x = 95\,cm$

Explanation:

(i) $x_{1}$ and $x_{2}$ represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:

$x = 115\,cm$

(ii) Maximum accelerations is reached at $x_{1}$ and $x_{2}$ .

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(iii) Greatest net forces exerted on the particle are reached at $x_{1}$ and $x_{2}$ .

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3 years ago

A 65kg person throws a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the

asambeis [7]

Answer:

v₁ = 2.48m/s, v₂ = 0.02m/s

Explanation:

Momentum p must be conserved. p = mv

1) First person throwing the snow ball. The momentum before the throw:

p = (65kg + 0.045kg) * 2.5 m/s

The momentum after the throw:

p = 65kg * v₁ + 0.045kg * 30m/s

Solving for the velocity v₁ of person 1:

v₁ = ((65kg + 0.045kg) * 2.5 m/s - 0.045kg * 30m/s) / 65kg = 2.48m/s

2) Second person catching the ball. The momentum before the catch:

p = 0.045kg * 30m/s + 60kg * 0m/s

The momentum after the catch:

p = (60kg + 0.045kg) * v₂

Solving for velocity v₂ of person 2:

v₂ = 0.045kg * 30m/s / (60kg + 0.045kg) = 0.02 m/s

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3 years ago