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Lana71 [14]
3 years ago
7

In the diagram of the swimmer, which force is greater, buoyancy or weight, and how do we know?

Physics
1 answer:
Anettt [7]3 years ago
6 0
Buoyancy and weight are pretty much equal however there is slightly more buoyancy since the swimmer isn't sinking and is still floating on the surface
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Reflection and refraction 


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4 years ago
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A particle with charge q = +5e and mass m = 8.2×10-26 kg is injected horizontally with speed 1.1×106 m/s into the region between
loris [4]

Answer:

d = 1.27m

Explanation:

Given m = 8.2×10-26kg, v = 1.1×10⁶m/s, q = +5e = 5×1.6×10‐¹⁹ C.

E = 49kN/C = 49000N/C

The displacement is given by

d = 1/2× mv²/qE = 1/2 × 8.2×10-²⁶ × (1.1×10⁶)²/(5×1.6×10-¹⁹ ×49000) = 1.27m

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3 years ago
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A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi
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Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

mass of the small sphere, m = 2.30 Kg

when the Point is outside the sphere the Force between them is equal to

      F = \dfrac{GMm}{r^2}   when r>R

When Point is inside the Sphere

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}  when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

       F = \dfrac{GMm}{r^2}[/tex]

       F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}

               F = 1.082 x 10⁻⁸ N

b) r = 2.65 m

      F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}

      F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}

     F = 5.68\times 10^{-9}\ N

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3 years ago
Which of the following are located outside the nucleus of an atom? Select all that
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Electrons are located outside the nucleus
3 0
3 years ago
A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1
ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) =  1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a  c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0
3 years ago
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