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Tcecarenko [31]

2 years ago

6

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in this collisi

on? Explain why or why not.

2 answers:

mart [117]2 years ago

6 0

Conservation of momentum<span> is a fundamental law of physics. This law states that the </span>momentum<span> of a system is constant if there are </span>no external forces acting on the system. In a situation in which two balls, each with a mass of 0.5 kg, collide on a pool table<span> the law of conservation of momentum is not satisfied because there are external forces that moved the balls. </span>

swat322 years ago

4 0

Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s, and the total momentum after the collision is
<span>1.5 kg • m/s. The momentum stays the same after the collision.</span>

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Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0

1 year ago

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

Ad libitum [116K]

Answer:

$F=(3i+3.6j)\ N$

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, $u=(1i+0j)\ m/s$

After 8 seconds, final velocity of the puck, $v=(6i+6j)\ m/s$

Let the x and y component of force is given by $F_x\ and\ F_y$ .

x component of force is given by :

$F_x=m\times \dfrac{v-u}{t}$

$F_x=4.8\times \dfrac{6-1}{8}$

$F_x=3\ N$

y component of force is given by :

$F_y=m\times \dfrac{v-u}{t}$

$F_y=4.8\times \dfrac{6-0}{8}$

$F_y=3.6\ N$

So, the component of the force is $F=(3i+3.6j)\ N$ . Hence, this is the required solution.

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2 years ago

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.500 mm. This results in an interference pat

ipn [44]

Answer:

0.000109375 m

Explanation:

d = Distance between grating = 0.5 mm

m = Order

$\lambda_1=\dfrac{d}{8}$

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$m-\dfrac{1}{2}$

For fourth order minima

$dsin\theta=(4-\dfrac{1}{2})\lambda_1$

For second maxima

$dsin\theta=2\lambda_2$

From the two equations we get

$(4-\dfrac{1}{2})\lambda_1=2\lambda_2\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\lambda_1}{2}\\\Rightarrow \lambda_2=\dfrac{(4-\dfrac{1}{2})\dfrac{0.5\times 10^{-3}}{8}}{2}\\\Rightarrow \lambda_2=0.000109375\ m$

The wavelength is 0.000109375 m

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2 years ago

The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a

Pie

Explanation:

It is given that,

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Velocity, v = 11 m/s

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Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

$J=F.\Delta t$

$J=10^6\ N\times 20\ s$

$J=2\times 10^7\ N-s$

(b) Impulse is also given by the change in momentum i.e.

$J=\Delta p=p_f-p_i$

$J=p_f-p_i$

$p_f=J+p_i$

$p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s$

$p_f=20143000\ kg-m/s$

(c) For new velocity,

$v_f=\dfrac{p_f}{m}$

$v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}$

$v_f=1549.46\ m/s$

Hence, this is the required solution.

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To begin to lift the car, a hydraulic jack must produce <em>more than 9,800N</em> of force.

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2 years ago

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