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2 years ago

Two difference between thrust and upthrust .

1 answer:

6 0

Upthrust is the upward force exerted by fluids on the surface of an object immersed in fluids. Thrust:- It is the force acting perpendicular to the surface. Upthrust:- It is the upward force exerted by the fluid on the surface of an object immersed in liquid.

I hope it's help you

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3. When two liquids are mixed and a solid

sashaice [31]

Answer:

Explanation:

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2 years ago

Using Planck’s constant as h = 6.63 E-34 J*s, what is the wavelength of a proton with a speed of 5.00 E6 m/s? The mass of a prot

Marrrta [24]

De Broglie's identity gives the relationship between the momentum and the wavelength of a particle:
$p=mv= \frac{h}{\lambda}$
where
p is the particle momentum
m is its mass
v its velocity
h is the Planck constant
$\lambda$ is the wavelength

By re-arranging the equation, we get
$\lambda= \frac{h}{mv}$
and by using the data about the proton, given in the text, we can find the proton's wavelength:
$\lambda= \frac{h}{mv} = \frac{6.63 \cdot 10^{-34} Js}{(1.66 \cdot 10^{-27} kg)(5.00 \cdot 10^6 m/s)} =7.99 \cdot 10^{-14} m$

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3 years ago

The SAF operates the M113 Ultra APC.

HACTEHA [7]

The volume of the object must be no larger than $11.15 m^3$ .

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

$\rho = 1000 kg/m^3$

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

$\rho = \frac{m}{V}$

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

$m=1.115\cdot 10^4 kg$

Therefore, we can re-arrange the equation to find its volume:

$V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3$

So, the volume of the object must be no larger than $11.15 m^3$ .

Learn more about density:

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brainly.com/question/8441651

#LearnwithBrainly

8 0

3 years ago

Visit this website and locate the element mercury, whose chemical symbol is Hg. Click on the square to read about the history, p

Irina-Kira [14]

We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.

6 0

2 years ago

Read 2 more answers

Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a

raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0

2 years ago