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2 years ago

The nuclear particle carrying a charge of +1e is the

Physics

2 answers:

babunello [35]2 years ago

6 0

The nuclear particle is a proton

aalyn [17]2 years ago

6 0

Inside the nucleus only two particles are present

1. Neutrons

2. Protons

These are also known as nucleons as combined

neutrons are chargeless and having mass same as that of protons while protons are positively charged particles and its charge is same in magnitude as that of electrons

SO the charge given here is +1e which means the magnitude of charge is same as the magnitude of electron but it is a positive charge

<u>so it must be proton</u>

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As a marble rolls across the floor, it gradually slows to a stop due to friction. Which statement best describes the change in m

Olegator [25]

Answer:

I belive it would be "C"

Explanation:

If it was any of the other answers "B" it would instantly stop. "A" it would roll forever.

3 0

2 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

2 years ago

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

$= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

3 0

3 years ago

A flare is shot from a snowmobile with an initial velocity of 4.3 m/s at 90*, while the snowmobile is moving at 8.5 m/s at 0.0*.

Sedaia [141]

The driver is tooling along in his snowmobile, pointed north,
at 8.5 m/s.

He's carrying the flares with him, so the flares are also moving north
at 8.5 m/s.

When he fires the flare straight up, it has a vertical velocity of 4.3 m/s
straight up, and a horizontal velocity of 8.5 m/s towards the north.

The magnitude of the net velocity is √(4.3² + 8.5²) .
That's about 9.53 m/s, at some angle between straight up
and straight north.

The angle above horizontal is the angle that has a tangent of 4.3/8.5 .
I'll let you work out the angle.

8 0

2 years ago

A dog runs to chase a ball. He travels 6m in 3.2 seconds. How fast is the dog running?

Artyom0805 [142]

So v=d/s so the answer is 6/3.2 so the answer is 1.87m/s

4 0

2 years ago