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2 years ago

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there are 3 balls , ball 1 has a mass of 2 kg ,ball number 2 has a mass of 0.5 kg, and ball number 3 has the mass of 0.02 kg ...

for all the balls to have the same momentum which one has the greatest velocity?

1 answer:

viva [34]2 years ago

7 0

Answer:

yes

Explanation:

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Time = 60 minutes

2) Constant angular velocity => (4π / 60) = (x / 20) => x = 4π / 3 inches

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Answer: 4.19 inches

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A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o

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Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

a)

Initial energy

$KE_i=\dfrac{1}{2}mu^2$

$KE_i=\dfrac{1}{2}0.3\times 10^2$

KEi= 15 J

Final kinetic energy

$KE_f=\dfrac{1}{2}mv^2$

$KE_f=\dfrac{1}{2}0.3\times 5^2$

KEf=3.75 J

The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

b)

The minimum value to complete the circular arc

$V=\sqrt{r.g}$

Now by putting the values

$V=\sqrt{0.8\times 10}$

V= 2.82 m/s

So kinetic energy KE

$KE=\dfrac{1}{2}mV^2$

$KE=\dfrac{1}{2}0.3\times 2.82^2$

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

= 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

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A certain 60.0 Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 11.6 kV/m.

yuradex [85]

Explanation:

Given that,

Frequency of the power line, f = 6 Hz

Value of maximum electric field strength of 11.6 kV/m

(a) The wavelength of this very low frequency electromagnetic wave is given by using relation as :

$c=f\lambda$

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8\ m/s}{60\ Hz}$

$\lambda=5\times 10^6\ m$

(b) As its can be seen that the wavelength of this wave is very high. It shows that it is a radio wave.

(c) The relation between the maximum magnetic field strength and maximum electric field strength is given by :

$B_0=\dfrac{E_0}{c}\\\\B_0=\dfrac{11.6\times 10^3}{3\times 10^8}\\\\B_0=3.86\times 10^{-5}\ T$

So, the maximum magnetic field strength is $3.86\times 10^{-5}\ T$ .

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In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i

pantera1 [17]

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

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d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

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