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guajiro [1.7K]
3 years ago
8

An isotope has the same ________, but a different ________ .

Physics
2 answers:
11111nata11111 [884]3 years ago
6 0
I<span>sotopes of an element have the same number of protons but different number of neutrons.</span>
Lunna [17]3 years ago
4 0
An isotope is one of the forms of an element that has the same number of protons, but a different number of neutrons.
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Jane has a mass of 55 kg and his body covers 700 nails with a surface area of 1.00 mm,
Oksana_A [137]

We have,

  • Jane mass is 55 kg
  • His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000

We know that,

  • Pressure = Force/Area

Let's calculate force as we already have area;

  • F = ma
  • F = 55 × 9.8 { Acceleration due to gravity }
  • F = 539 N

Now, if should she would be on 700 nails then pressure will be;

  • P = F/A
  • P = 539/7 × 10000
  • P = 5390000/7
  • P = 770,000 Pascal

And if should would be on a 1 nail only,

  • P = F/A
  • P = 539/1 × 1000000
  • P = 539000000 Pascal

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6 0
3 years ago
PLEASE HELP ASAP!!!<br> GIVING BRAINLIEST!!!<br> 40 points!!
Nana76 [90]
W = Fd = 4(2100) = 8400 J

So the answer is A) 8400 J


I was just rewriting my notes on the work lesson I did in class today, so I saw this question at the perfect time!! :)

Hope it helps!! :)
4 0
3 years ago
Read 2 more answers
A child is swinging back and forth with a constant period and amplitude. Somewhere in front of the child, a stationary horn is e
Amanda [17]

Answer:

Explanation:

  We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .

f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .

f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source  while swinging to and from before the source of sound .

3 0
3 years ago
The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?
vitfil [10]

The expression of the electric flux is

\Phi = \frac{Q}{\epsilon_0}

Here,

Q = Total charge enclosed in the closed surface

\epsilon_0 = Permittivity due to free space

Rearranging to find the charge,

Q = \epsilon_0 \Phi

Replacing with our values we have finally

Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)

Q = 1.6284*10^{-8} C (\frac{10^9nC}{1C})

Q = 0.1684nC

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0
3 years ago
If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
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