Ask question

Ask question

All categories

2 years ago

6

One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its hi

ghest point, comes to within 10 cm of the end of the ramp without going off. You give the puck a push, releasing it with a speed of 5.0 m/s when it is 8.50 m from the end of the ramp. The puck's speed after traveling 3.0 m is 4.0 m/s.

1 answer:

Aneli [31]2 years ago

4 0

Answer:

Explanation:

Given

initial speed(u)=5 m/s

Final speed(v)=4 m/s

Distance traveled=3 m

using equation of motion

$v^2-u^2=2as$

$4^2-5^2=2(a)(3)$

$a=\frac{-3}{2}=-1.5 m/s^2$

after this its final velocity will be zero

$v^2-u^2=2as$

$0^2-4^2=2\times (-1.5)\times s$

s=5.33 m

Total distance=3+5.33=8.33 m

Thus he will not be able to win the game

You might be interested in

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

1 year ago

A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h

Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0

2 years ago

In the picture below, explain why the bear fell. Use FRICTION to explain your answer.

algol13

The bear fell because it slides to the surface of ice due to lack of friction.

One of these theories is that friction<span> causes the liquid layer of water to form on </span>ice<span>. </span>Friction<span> is the force that generates heat whenever two objects slide against each other. If you rub your hands together, you can feel them heat up. That's </span>friction<span> at work. When a </span>skate<span> moves over the surface of </span>ice, the friction<span> between the </span>skate<span> and the </span>ice<span> generates heat that melts the </span>outermost<span> layer of </span>ice<span>.</span>

5 0

2 years ago

A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface

valentinak56 [21]

Answer:

$\frac{kQ}{r^2} r^$

Explanation:

Electric field strength= Force/unit charge

E= (kQq/r²)/q ₓ r

where r is the unit vector in the direction of unit charge

E= $\frac{kQ}{r^2} r^$

4 0

2 years ago

the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

vivado [14]

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

3 0

2 years ago