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3 years ago

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A child is bored in the back of the car and starts to time how long the car takes to pass some 100m markers on the side of the r

oad. Initially it takes 5 seconds to pass between them. The car accelerates at 4m/s2 for 2 seconds.
Calculate the initial velocity of the car before the acceleration.

1 answer:

Delvig [45]3 years ago

4 0

Answer:

Follow my insatagram onlypiccolo for answer

Explanation:

ok?

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A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from

svlad2 [7]

Answer:

) the uniform disk has a lower moment of inertia and arrives first.

Explanation:

(a) the uniform disk has a lower moment of inertia and arrives first.

(b) Let's say the disk has mass m and radius r, and

the hoop has mass M and radius R.

disk: initial E = PE = mgh

I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh

m cancels, leaving v² = 4gh / 3

hoop: initial E = Mgh

I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh

M cancels, leaving V² = gh

Vdisk = √(4gh/3) > Vhoop = √(gh)

3 0

3 years ago

As more and more bulbs are connected in series to a flashlight battery, what happens to the brightness of each bulb? Assuming th

balu736 [363]

Answer:

$P_1 = P_2 = \frac{P}{2}$

so each bulb brightness becomes half of its given or indicated power

$P_1 = P_2 = \frac{V^2}{R}$

so both bulb will glow same power as indicated

Explanation:

Let the indicated power on the bulbs is given as P and its rated voltage is V

so here resistance of each bulb is given as

$R = \frac{V^2}{P}$

now if the two bulbs are connected in series so we will have

$R_{eq} = R_1 + R_2$

$R_{eq} = 2\frac{V^2}{P}$

now the current in the circuit is given as

$i = \frac{V}{R_{eq}$

$i = \frac{P}{2V}$

now brightness of each bulb is given as

$P_1 = P_2 = i^2 R$

$P_1 = P_2 = \frac{P}{2}$

so each bulb brightness becomes half of its given or indicated power

Now if the two bulbs are connected in parallel

then the net voltage across each bulb is "V"

so we will have

$P_1 = P_2 = \frac{V^2}{R}$

so both bulb will glow same power as indicated

6 0

4 years ago

what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I

denis23 [38]

Answer: hello some part of your question is missing attached below is the missing detail

answer :

<em>w</em>f = M( v cos∅ )D / I

Explanation:

The Angular speed <em>wf </em>of the system after collision in terms of the system parameters and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) <em>w</em>f

where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )

In terms of system parameters and I

<em>w</em>f = M( v cos∅ )D / I

5 0

3 years ago

Celina has a water sample that's contaminated with salt and microorganisms. Which method should she use to purify the water.

sergiy2304 [10]

She should filter the water with carbon and oxygen

7 0

3 years ago

Read 2 more answers

Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a

worty [1.4K]

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$ . Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$ .

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ .

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$ .

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$ .

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$ .

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$ , would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$ .

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$ . The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$ .

6 0

2 years ago