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3 years ago

Without understanding mathematics, it is practically impossible to see its applicability and beauty

Physics

1 answer:

Kamila [148]3 years ago

4 0

True

Explanation:

In order to apply and experience the beauty of mathematics, a good comprehension of the discipline is required.

In fact, this is general to anything in life. To fully maximize any potential, one must be well groomed about what exactly that thing is.

Mathematics is a science that deals with the logic of shapes and numbers. There are different branches of this discipline with real world adoption.

The tenets of mathematics cuts across business, science, arts, e.t.c.

Learn more:

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What is the magnitude of the orbital velocity of the earth in m/s?

Lana71 [14]

Circumference C=2πr
C=2π(1.5x10^8)=9.42x10^8 

In 365 Days there are 8760hr 

V=distance/time 

V=(9.42x10^8)/8760=107534.2km/hr

8 0

3 years ago

What is the frictional force between a box and the floor it is being pulled across if, the kinetic coefficient of friction is 0.

Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

n = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

f = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0

3 years ago

P14.003A spherical gas-storage tank with an inside diameter of 8.1 m is being constructed to store gas under an internal pressur

Artist 52 [7]

Answer:

The minimum wall thickness required for the spherical tank is 0.0189 m

Explanation:

Given data:

d = inside diameter = 8.1 m

P = internal pressure = 1.26 MPa

σ = 270 MPa

factor of safety = 2

Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?

The allow factor of safety:

$\sigma _{a} =\frac{\sigma }{factor-of-safety} =\frac{270}{2} =135MPa$

The minimun wall thickness:

$t=\frac{Pd}{4\sigma _{a} } =\frac{1.26*8.1}{4*135} =0.0189m$

3 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

4 years ago

लाइ फ्युजका 4 Why MCB is called the developed form of fuse? Give the reason

DerKrebs [107]

Miniature circuit breakers is called the developed form of fuse because MCBs are more sensitive to current than fuses. They immediately detect any abnormality and switch off the electrical circuit automatically. This prevents any permanent damage to electrical appliances and human beings

8 0

3 years ago