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2 years ago

2

Physics

1 answer:

Xelga [282]2 years ago

8 0

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where U is potential energy and K is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

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Explanation:

Given parameters:

Distance hopped = 84m

Displacement = 84m due east

Time = 7s

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Speed of kangaroo = ?

Velocity of kangaroo = ?

Solution:

To solve this problem,

Speed = $\frac{distance}{time }$ = $\frac{84}{7}$ = 12m/s

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3 years ago

Physics. Need help. Brainlieast answer for most/ all of the answers answered

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ALL of the following work assumes NO AIR RESISTANCE:

1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

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4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion

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8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s; b). 29.7 m/s

10). 7.24 meters

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3 years ago

un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l

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Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

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3 years ago

If two cars are 5 m apart, and one car has a mass of 2,565 kg and the other

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Answer:

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3 years ago

Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay

kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

$mu=(m+M)v$

where

m = 0.200 kg is the mass of the koala bear

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M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

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