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2 years ago

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A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene

rgy just as they leave the ground. Air resistance is negligible. Which of the following statements about these frogs are correct?
The heavier frog goes higher than the lighter frog
The heavier frog goes higher than the lighter frog
They collide and stick together.
the lighter frog is moving faster than the heavier frog.

1 answer:

Ilia_Sergeevich [38]2 years ago

5 0

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

$K(light) = K(heavy) = \frac{1}{2} *ml*vol^{2} = \frac{1}{2}*mh*voh^{2}$

Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax$

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

$hmax =\frac{vo^{2} }{2*g}$

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

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A man climbs a ladder. Which two quantities can be used to calculate the energy stored of the man at the top of the ladder.

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Answer:The answer must be The weight of the man and the vertical distance moved.

Explanation: you calculate it by the force you applied times the distance you moved

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1 year ago

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

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To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

$\omega =$ Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

$1000J > \frac{1}{2}mv^2$

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2$

$1000J = \frac{1}{2}mv^2$

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

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8. Rubbing a plastic bag and a balloon with a cloth gives both objects a net negative charge. The balloon's

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Answer:

0.214 m

Explanation:

In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.

Therefore, we can write:

$k\frac{q_1 q_2}{r^2}=mg$

where

k is the Coulomb constant

$q_1=-1\cdot 10^{-10}C$ is the charge on the balloon

$q_2=-1\cdot 10^{-5} C$ is the charge on the bag

r is the separation betwen the bag and the balloon

$m=0.02 g=2\cdot 10^{-5} kg$ is the mass of the bag

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for r, we find the distance at which the bag must be held:

$r=\sqrt{\frac{kq_1 q_2}{mg}}=\sqrt{\frac{(9\cdot 10^9)(-1\cdot 10^{-10})(-1\cdot 10^{-5})}{(2\cdot 10^{-5})(9.8)}}=0.214 m$

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2 years ago

Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman

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Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2$

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

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2 years ago

A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m

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<span>(a)

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get:

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component.

(b)
To much the same end do we derive the vertical component:

18.0*sin37.5 = v_y = 10.96 ms^-1

Which we then divide by acceleration, a_y, to derive the time till maximal displacement,

10.96/9.8 = 1.12 s

Finally, doubling this value should yield the particle's total time with r_y > 0

<span>2.24 s

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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2 years ago