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3 years ago

What is the volume of a bar of soap that is 9 cm long, 5 cm wide, and 2 cm high?

Physics

2 answers:

GalinKa [24]3 years ago

7 0

Answer:

90cm

Explanation:

2x5=10

10x9=90

Nataly_w [17]3 years ago

3 0

Answer:

90 cm

Explanation:

To find volume, the formula is V=LWH

L is the length, W is the width and H is the height:

9 x 5 x 2 = 90

So V = 90 cm

Hope this helps you out! : )

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Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

kodGreya [7K]

Answer:

$v = 4.048 *10^6$ m/s

Angular frequency = $1.92 * 10^7$

Explanation:

As we know

$v = \frac{qBr}{m}$

q is the charge on the electron = $3.2 * 10^{-19}$ C

B is the magnetic field in Tesla = $0.4$ T

r is the radius of the circle = $0.21$ m

mass of the electrons = $6.64 * 10^{-27}$ Kg

Substituting the given values in above equation, we get -

$v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6$ m/s

Angular frequency =

$\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7$

8 0

2 years ago

When an exothermic reaction releases thermal energy, this energy is usually_____.

saw5 [17]

Explanation:

A chemical reaction in which heat or energy is released is known as an exothermic reaction.

On the other hand, when two objects are placed together and heat flows from hotter object to colder object then this process is known as conduction. Therefore, energy is dissipated in conduction process.

Since energy released released goes into the atmosphere and is not used anywhere.

Thus, we can conclude that when an exothermic reaction releases thermal energy, this energy is usually not useable to do work and it is dissipated by conduction.

8 0

3 years ago

The record of an earthquake obtained from a seismic instrument is a(n) ________.

serious [3.7K]

<span>The record of an earthquake obtained from a seismic instrument is a </span>
SEISMOGRAM

6 0

3 years ago

A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. How much energy is lost? Group of answer choices 0.5

bogdanovich [222]

Answer: A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. Energy loss will be 0.29J

Explanation: To find the correct answer, we have to know more about the Gravitational potential energy.

<h3>What is gravitational potential energy?</h3>

The energy possessed by a body by virtue of its position in gravitational field of earth is called gravitational potential energy.
The gravitational potential energy of a body at a height h with respect to the height h will be,

$U=mgh$