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Vera_Pavlovna [14]
2 years ago
8

A beaker is filled to the brim with water. A solid object of mass 3.00 kg is lowered into the beaker so the object is fully subm

erged in the water (see the drawing). During this process, 2.00 kg of water flows over the rim and out of the beaker. What is the buoyant force that acts on the submerged object, and, when released, does the object rise, sink, or remain in place?
Physics
1 answer:
Sholpan [36]2 years ago
6 0

Answer:

Bouyant Force is 19.6N and the object will sink

Explanation:

The buoyant force is equal to the weight of displaced water(Archimedes Principle). Since 2.00 kg of water were displaced, the buoyant force would be equal to:

Fb = Fw = mg = (2.00 kg)(9.81 m/s²) = 19.6 N

Where Fb is buoyant force

Where Fw is weight of displaced water

The object will float if the buoyant force is greater than its weight, and sink if the reverse is true. Since the object's mass is 3.00 kg, its weight is:

Fw = mg = (3.00 kg)(9.81 m/s²) = 29.4 N

Since the object's weight, 29.4 N, is greater than the weight of displaced water, the object will sink.

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Light from a lamp is shining on a surface. how can you increase the intensity of the light on the surface?
vlada-n [284]
The intensity of a light in a surface follows the inverse square law formula which can be mathematically expressed as,
                         I = k/d²
where I is intensity, d is distance, and k is the proportionality constant. For us to increase the intensity, we should lower the distance from the source to the surface. 
4 0
1 year ago
Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f
Lunna [17]

Answer:

The final speed will be the same for the children on the shorter side and on the longer side.

Explanation:

This is because since the they are the same distance above the ground, their potential energy which is a function of mass, acceleration due to gravity and vertical height are the same.

PE = Mass × gravity × vertical height

At this point, we can deduce that the horizontal length of the slide has no effect on the potential energy. Only the vertical height does.

All this potential energy is converted to kinetic energy at the end of the slide. Since the potential energy is the same, then the kinetic energy will be the same and thus their velocity is the same.

Mathematically, consider that PE = mgh and KE = \frac{1}{2}mv^{2}

at the bottom of the slide, since energy has to be conserved, PE must be equal to KE.

mgh = \frac{1}{2}mv^{2}

final velocity of the child , v = \sqrt{2gh}

It shows the final velocity is only a function f acceleration due to gravity and height.

Thus, making their velocities equal.

8 0
1 year ago
A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Bro
aleksley [76]

Answer:

Potential difference though which the electron was accelerated is 2.67\times 10^{-6}\ uV\  .

Explanation:

Given :

De Broglie wavelength , \lambda=750\ nm.

Plank's constant , h=6.626\times 10^{-34}\ J.s \ .

Charge of electron , e=-1.6\times 10^{-19}\ C.

Mass of electron , m=9.11\times 10^{-31}\ kg.m=9.11\times 10^{-31}\ kg.

We know , according to de broglie equation :

\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}

Putting all values in above equation we get ,

V=2.67\times 10^{-6}\ uV .

Hence , this is the required solution.

5 0
2 years ago
two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f
Zina [86]

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

5 0
2 years ago
A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit
Maslowich

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= -  GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = \frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}

=  - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}

=   - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

8 0
1 year ago
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