1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bezimeni [28]
3 years ago
8

Wooden ball lies in a vessel with water so that half of it is immersed in water.

Physics
1 answer:
Ugo [173]3 years ago
5 0

Downward force acting on the ball is 19.6N

Net force acting on the ball is 1960V N

<u>Explanation:</u>

<u />

Given:

Mass of the ball, m = 2kg

Density of ball, σ = 800 kg/m³

Density of water, ρ = 1000 kg/m³

Downward force acting by the ball in the vessel = mg

where, g = 9.8m/s²

F = 2 X 9.8

F = 19.6N

Net force acting on the ball:

Fnet = (ρ - σ) Vg

where,

V is the volume of water

Fnet = (1000 - 800) V X 9.8

Fnet = 1960V N

If the volume is known, then substitute the value of V to find the net force.

Thus, Downward force acting on the ball is 19.6N

         Net force acting on the ball is 1960V N

You might be interested in
What is the formula for force?
Vitek1552 [10]
Choice 'b' is one possible way to state
Newton's second law of motion.

The other choices are meaningless.
8 0
3 years ago
platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin
posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

a=g=9.8 m/s^2 is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s

For the diver jumping from 10 m, s = 10 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

s=ut+\frac{1}{2}at^2

And since u = 0, it can be reduced to

s=\frac{1}{2}at^2

For the diver jumping from 5 m, s = 5 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s

For the diver jumping from 10 m, s = 10 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s

5 0
3 years ago
In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text
Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

 gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second ,  change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

=  4.859 x 10⁻² oscillations .

5 0
3 years ago
Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod
cluponka [151]

Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

Explanation:

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

F=\dfrac{mv^2}{r}

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

F'=\dfrac{2m\times (4v)^2}{4r}\\\\F'=\dfrac{2m\times 16v^2}{4r}\\\\F'=8\dfrac{mv^2}{r}\\\\F'=8F

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

8 0
3 years ago
A person is standing on a raft; their
krok68 [10]

Answer:

The volume of water displaced by the raft is 0.233 m³

Explanation:

The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid

The given parameters are;

The combined mass of the person and the raft, m = 233 kg

The liquid on which the raft is located = Water

The density of water, \rho _{water} = 1000 kg/m³

Weight = Mass, m × g

Where;

m = The mass of the object

g = The acceleration due to gravity = 9.8 m/s²

Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg

The combined weight of the person and the raft, W_{combined} = 233 kg × 9.8 m/s² = 2,283.4 N

Therefore;

The buoyant force = 2,283.4 N = The weight of the water displaced

The mass of the water displaced, m_{water}, = 2,283.4 N/(9.8 m/s²) = 233 kg

Density = Mass/Volume

The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.

3 0
2 years ago
Read 2 more answers
Other questions:
  • 3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,
    11·1 answer
  • A hiker walks 2.00 km north and then 3.00 km east, all in 2.50 hours. Calculate the magnitude and direction of the hiker’s (a) d
    10·1 answer
  • 12. What does positive and negative acceleration indicate?
    6·1 answer
  • Are we actually touching nothing but electrons in reality or can we actually feel things without the electron barier?
    10·1 answer
  • Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe
    7·1 answer
  • Which of the following are transferred or shared when two atoms react chemically? *
    9·1 answer
  • To catch a fast-moving ball, you extend your hand forward before contact with the ball and let it ride backward in the direction
    8·1 answer
  • Please help me answer this it is due by Sunday night.
    12·1 answer
  • Two identical light bulbs, each with resistance R = 2  are connected to a source with E = 8 V and negligible internal resistanc
    10·1 answer
  • 1. Describe the three Newton's Law of Motion.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!