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musickatia [10]

3 years ago

14

The driver of a car traveling at 30.5 m/s slams on the brakes so that the car undergoes a constant acceleration, skidding to a c

omplete stop in 4.5 s. What is his average acceleration?
-2.6 m/s^2
2.6 m/s^2
6.8 m/s^2
-6.8 m/s^2

1 answer:

dolphi86 [110]3 years ago

4 0

Answer:

-6.8 m/s²

Explanation:

Given:

v₀ = 30.5 m/s

v = 0 m/s

t = 4.5 s

Find: a

a = (v − v₀) / t

a = (0 m/s − 30.5 m/s) / 4.5 s

a = -6.8 m/s²

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You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of t

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Answer:

Height with sound ignored = Gravity x Time taken = 9.8 x 8.60 = 84.28 meters

Time taken by the sound = 84.28/330 = 0.255 seconds

Height with sound involved = (84.28 x 0.255) + 84.28 = 105.80 meters

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You have a ball with a mass 1.3 kg tied to a rope, and you spin it in a circle of radius 1.8m. If the ball is moving at a speed

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Answer:

$6.5\; {\rm N}$ .

Explanation:

When an object travel at a speed of $v$ in a circle of radius $r$ , the (centripetal) acceleration of that object would be $a = (v^{2} / r)$ .

In this question, the ball is travelling at $v = 3\; {\rm m\cdot s^{-1}}$ in a circle of radius $r = 1.8\; {\rm m}$ . The (centripetal) acceleration of this ball would be:

$\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}$ .

By Newton's Laws of Motion, for an object of mass $m$ , if the acceleration of that object is $a$ , the net force on that object would be $m\, a$ . Since the acceleration of this ball is $a = 5\; {\rm m\cdot s^{-2}}$ , the net force on this ball would be:

$\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}$ .

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Answer:

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3 years ago

Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a

mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where, taking upward as positive direction:

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the pebble is launched horizontally)

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

$v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s$ (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

$v_x = 20.0 m/s$

So, the final speed of the pebble as it strikes the ground is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s$

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

$u_y = 20.0 m/s$

So we can find the final vertical velocity using the same suvat equation as before:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

$u_y = -20.0 m/s$

Using again the same suvat equation:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0

3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago