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3 years ago

State archimedes principle

Physics

1 answer:

Travka [436]3 years ago

5 0

Answer:

Kindly check explanation

Explanation:

Archimedes principle is a very important concept in fluid mechanics which is credited to the Greek mathematician called Archimedes. The law simply states that; When an object is fully or partially immersed in a fluid , the immersed body experiences an upthrust(Weight loss) which is equal to the weight of fluid displaced by the immersed object.

In essence, the upthrust or volume of Displacement experienced by the object will be equal to the volume of the the object which is immersed in the fluid. Such that the force of buoyancy matches the weight of object immersed in the fluid.

One of the most common illustrations of Archimedes is in the navigation of ships whose upthrust increases as the weight of the ship increases such that the upthrust force balances the weight of the ship in water.

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A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy a

vovikov84 [41]

This question involves the concept of kinetic energy.

The student's claim is "right".

<h3>Kinetic Energy</h3>

The energy possessed by a body, by the virtue of its motion is called kinetic energy. Mathematically it is given by the following formula:

$K.E =\frac{1}{2}mv^2$

where,

K.E = Kinetic energy
m = mass
v = velocity

Therefore,

For the paintball:

$K.E = \frac{1}{2}(4\ g)(90\ m/s)^2$

K.E = 16200 J

For the pellet:

$K.E = \frac{1}{2}(1\ g)(180\ m/s)^2$

K.E = 16200 J

Hence, both paintball and pellet will have same kinetic energy. The student is right.

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brainly.com/question/12669551

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2 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$

To find Ny, we need to find the tension T.
For this, we can equate the net horizontal force.

$F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N$

Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

$N_y= (40*9.8)-(169.8*sin59)=246.4N$

Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

$N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N$

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

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4 0

2 years ago

Read 2 more answers

What does an Astronomer study? *

ratelena [41]

Answer:

an astronomer studies planets, stars, moons, etc, or objects that are outside the field of Earth.

3 0

3 years ago

(b) The distance of mass from mass A if there is no gravitational force acted on C

shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

$F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}$

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ = 6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ = 6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

$F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}$

By plugging in the values and removing like terms, we get;

$\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}} = \dfrac{2 \times 0.05}{x^2}$

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0

3 years ago

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14

Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

$\omega_f^2 - \omega_i ^2 =2 \alpha \theta$

where

$\omega_f = 3.14\cdot 10^4 rad/s$ is the final angular speed

$\omega_i = 1.10 \cdot 10^4 rad/s$ is the initial angular speed

$\theta= 2.00 \cdot 10^4 rad$ is the angular distance covered

$\alpha$ is the angular acceleration

Re-arranging the formula, we can find $\alpha$ :

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2$

Now we want to know the time the bit takes starting from rest to reach a speed of $\omega_f=7.85\cdot 10^4 rad/s$ . So, we can use the following equation: