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Klio2033 [76]
3 years ago
11

GIZMO

Physics
1 answer:
CaHeK987 [17]3 years ago
5 0
It is A or D but I believe A
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A piece of plastic has a net charge of +9.9 μC. How many more protons than electrons does this piece of plastic have? (e = 1.60
Leto [7]

Answer:

The number of protons 6.19 more than electron.

Explanation:

Given that,

Charge Q=+9.9\times10^{-6}\ C

We know that,

formula of charge

Q = nc

Where,

Q = total charge

n = number of protons

e = charge of electron

Put the value into the formula

n = \dfrac{Q}{e}

n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}

n=6.1875\times10^{13}\ C

According to statement of question

Divide the answer by 10^{13}

n=\dfrac{6.1875\times10^{13}}{10^{13}}

n=6.19\ C

Hence, The number of protons 6.19 more than electron.

6 0
3 years ago
A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.00 m abov
Mamont248 [21]

Answer:

radial acceleration is 41.8 m / s²

Explanation:

The acceleration for circular motion is

     a = v² / r

They also give us the X and Y position where the body falls when the rope breaks, let's write the projectile launch equations

     x = vox t

     y = v₀ₓ t - ½ g t2

Since the circle is horizontally the v₀ₓ is zero (v₀ₓ = 0)

     x = v₀ₓ t

     t = x / v₀ₓ

     y = - ½ g t²

Let's replace and calculate the initial velocity on the X axis

    y = - ½ g (x / vox)²

    v₀ₓ = √ (g x² / 2 y)

    v₀ₓ = √ [- (-9.8) 1.6² / (2 1.00)]

    v₀ₓ = 3.54 m / s

This is the horizontal velocity, but since it circle is in  horizontal position it is also the velocity of the body at the point of rupture.

Now we can calculate the radial acceleration

        a = v² / r

       a = 3.54² / 0.300

       a = 41.8 m / s²

5 0
3 years ago
What do you call the procedure that helps you determine the volume of an irregularly shaped object, while using a graduated cyli
Lena [83]
Water displacement. You fill a graduated cylinder with an amount of water, place the object inside the graduated cylinder, and then measure the new water level. The change in volume of the water is the volume of the object, assuming the object was completely submerged. 
8 0
3 years ago
A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the
iris [78.8K]

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

<h3>What's the expression of magnetic field inside a solenoid?</h3>
  • Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
  • n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
  • Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

  • So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

Learn more about the magnetic field inside the solenoid here:

brainly.com/question/22814970

#SPJ4

6 0
1 year ago
20.0 -kg cannonball is fired from a cannon with muzzle speed of 1000m/s at an angle of 37.0° with the horizontal. A second ball
Dovator [93]

The mechanical energy for the first and the second ball is

10 ^{7}  \: joules.

Mass of the first ball = 20 kg

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 37°

The maximum height reached by the first ball is,

=   \frac{ u {}^{2} _{1}sin {}^{2} θ}{2g}

=    \frac{ {1000}^{2} sin {}^{2}37°}{2 \times 9.8}

= 18478.69 \: m

The maximum height of the first cannonball is 17478.69 m.

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 90 °

=   \frac{ u {}^{2} _{2}sin {}^{2} θ}{2g}

=   \frac{ 1000{}^{2}sin^{2} 90°}{2 \times 9.8}[tex] = 51020.41 \: m

For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height

So, the total mechanical energy is,

= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]

= 20 \times 9.8 \times 18478.64  \times  \frac{20}{2} (1000 \: cos37 °)

= 10 ^{7}  The potential energy at the maximum height, = m _{2}gh

= 20 \times 9.8  \times 51020.41

= 10 ^{7} \:J

Therefore, the total mechanical energy for the first and the

\:second \:  cannonball \:  is  \: 10 ^{7}  \:joules.

To know about energy, refer to the below link:

brainly.com/question/1932868

#SPJ4

5 0
1 year ago
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