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4 years ago

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From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The init

ial speed of the pebble is 9.0 m/s. (a) What is the acceleration (magnitude and direction) of the pebble during the downward motion? Is the pebble decelerating? Explain, (b) After 0.50 s, how far beneath tine cliff t

1 answer:

In-s [12.5K]4 years ago

7 0

Answer:

a) the acceleration is -9.8 m/s^2 (gravity's acceleration), the magnitude is 9.8 m/s^2 and the direction is downwards.

The pebble is not decelerating since the magnitude of it's speed is increasing, even if it is on a negative direction.

b) after 0.5 s the pebble is 5.725 meters down. (or -5.725m from it's initial position.)

Explanation:

a) the acceleration is given only by gravity, it's the only force acting on the pebble, the slingshot only gives the pebble an initial speed.

b) To calculate the position we use the equation

y0 + v0·t + ½·g·t² = yf

replacing:

0 + (-9.0m/s) * 0.5s + 1/2 (-9.8)*0.5² = yf

-5.725 = yf

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Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

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y=2.5 m

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Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

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$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

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Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

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3 years ago

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor

Sveta_85 [38]

Answer:

d) 12 V

Explanation:

Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.

We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

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3 years ago

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Answer:

$\boxed{\sf Distance \ travelled = 72 \ m}$

Given:

Initial velocity (u) = 0 m/s (Starts from rest)

Final velocity (v) = 24 m/s

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To Find:

Distance travelled by car (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

$\boxed{ \bold{s = (\frac{v + u}{2} )t}}$

By substituting value of v, u & t in the equation we get:

$\sf \implies s = ( \frac{24 + 0}{2} ) \times 6 \\ \\ \sf \implies s = \frac{24}{2} \times 6 \\ \\ \sf \implies s = 12 \times 6 \\ \\ \sf \implies s = 72 \: m$

$\therefore$

Distance travelled by car (s) = 72 m

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