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3 years ago

What is the difference between Mass and Weight? Why do we use mass instead of weight?

2 answers:

4 0

Mass is a measure of how much force it will take to change that path. ... Weight, on the other hand, is a measure of the amount of downwards force that gravity exerts on an object. This force increases with the object's mass: the more inertia it has, the harder gravity pulls.

Vladimir79 [104]3 years ago

4 0

In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity.

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Magnetic levitation has been used to innovate which aspect of life? A. communication B. mapping C. transportation D. space explo

bekas [8.4K]

I think the correct answer among the choices listed above is option C. Magnetic levitation has been used to innovate transportation. This innovation is commonly known as maglev. It is a new transportation technology where noncontacting vehicles travel above a guideway by magnetic fields.

4 0

4 years ago

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A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi

Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

$\Delta x_1=0.50m-0.35m=0.15m$

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

$k\Delta x_1=m_1g$

Solving for the spring constant k, we get:

$k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}$

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

$k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m$

Finally, we sum this to the unstretched length to obtain the length of the spring:

$x_2=0.35m+0.088m=0.44m$

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0

4 years ago

A wire carries 3.7 A of current. A second wire is placed parallel to the first 8.0 cm away. What is the current flowing through

IgorC [24]

Answer:

The current in second wire is 5.0 A.

(B) is correct option.

Explanation:

Given that,

Current in first wire = 3.7 A

Distance = 8.0 cm

We need to calculate the magnetic field due to the current carrying wire

Using formula of magnetic field

$B=\dfrac{\mu_{0}I}{2\pi r}$

Where, I = current

r = distance

Put the value into the formula

For first wire

$B = \dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}$ ...(I)

For second wire,

The distance is 8-3.7 = 4.3 cm

$B' = \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}$ ...(II)

The magnetic field in both the wires,

From equation (I) and (II)

$\dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}= \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}$

$I'=\dfrac{3.7\times4.3\times10^{-2}}{3.4\times10^{-2}}$

$I'=4.68\ A\ approx = 5.0\ A$

Hence, The current in second wire is 5.0 A.

8 0

4 years ago

A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds

san4es73 [151]

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

$E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\$

=218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1) $E_{0,\pi }$

$K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}$

$u'=+-0.283c$

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

$u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c$

6 0

3 years ago

After the supernova in a star’s life cycle, what else is formed in addition to a nebula?

DaniilM [7]

After the supernova they either form into a Black Hole, Neutron Star or White Dwarf. Depending

4 0

3 years ago

Read 2 more answers