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3 years ago

6

A 2 kg ball is released from rest at the top of a track and

1 answer:

xz_007 [3.2K]3 years ago

7 0

A) The formula for kinetic energy is E = 1/2 mv^2, so the energy of the ball is 1/2 * 2 * 10^2 = 100J. b) Energy is always conserved, and so if no energy is lost to resistive forces then all 100J of kinetic energy came from its potential energy at the top of the track. c) The formula for potential energy is E = mgh, which we can rearrange for h = E/mg. We know the energy, the mass and the strength of gravity, so we can find h = 100 / (2*9.81) = 5.10m.

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Imagine you're standing on a skateboard and your friend pushes you to the left with a force of 100 N. Neglecting friction and ai

ArbitrLikvidat [17]

It might be pull at a force of 100 N. I might be wrong.

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3 years ago

In a computer-based experiment to study diffraction, the width of the central diffraction peak is 15.20 mm. The wavelength of th

pogonyaev

Answer:

$a = 5.7 \times 10^{-5} m$

Explanation:

As we know that position of first minimum on the either side of central maximum is given as

$a sin\theta = \lambda$

$\theta =sin^{-1} \frac{\lambda}{a}$

so the width of the central maximum is given as

$W = L (2\theta)$

so we have

$15.20 \times 10^{-3} = 0.68 \times 2(sin^{-1} \frac{\lambda}{a})$

so we have

$0.011 = sin^{-1} \frac{\lambda}{a}$

$0.011 = \frac{638 nm}{a}$

$a = 5.7 \times 10^{-5} m$

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4 years ago

A 3.0 \Omega Ω resistor is connected in parallel to a 6.0 \Omega Ω resistor, and the combination is connected in series with a 4

GarryVolchara [31]

Answer:

The power dissipated in the 3 Ω resistor is P= 5.3watts.

Explanation:

After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.

The resultating resistor is of Req=6Ω.

I= V/Req

I= 2A

the parallel resistors have a potential drop of Vparallel=4 volts.

I(3Ω) = Vparallel/R(3Ω)

I(3Ω)= 1.33A

P= I(3Ω)² * R(3Ω)

P= 5.3 Watts

7 0

4 years ago

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The

nordsb [41]

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

7 0

3 years ago

How is the Sun similar to the stars that can be seen at night?

prisoha [69]

Answer:

A

Explanation:

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3 years ago

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