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3 years ago

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A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frict

ional force of 2 N, find (a) its acceleration, and (b) its initial velocity.

1 answer:

julia-pushkina [17]3 years ago

7 0

B it’s Intail velocity

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Which planet is smaller than Earth’s core?

makvit [3.9K]

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The answer <em><u>is C. Mars</u></em>. Mars and Mercury are both smaller than Earth's core. Hope this helps you :)

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3 years ago

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When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a con

Yuri [45]

Answer: C. the rod gains mass and the fur loses mass.

Explanation:Atomic particles have mass. The electron has a mass that is approximately 1/1836 that of the proton and with exchange exchange of charge this is also factored in. The movement of effect described above is known as the triboelectic charging process—charging by friction—which results in a transfer of electrons between the two objects when they are rubbed together. Plastic having a much greater affinity for electrons than animal fur pulls electrons from the atoms of fur, leaving both objects with an imbalance of charge. The plastic rod would have an excess of electrons and the fur has a shortage of electrons. Having an excess of electrons, the plastic is charged negatively and has more mass. In the same vein, the shortage of electrons on the fur leaves it with a positive charge and consequently with lesser mass.

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3 years ago

An air conditioner running with R-134a on a cycle executed under the saturationdome between the pressure limits of 0.8 MPa and 0

ohaa [14]

Answer:

The COP of the system is = 4.6

Explanation:

Given data

Higher pressure = 1.8 M pa

Lower pressure = 0.12 M pa

Now we have to find out high & ow temperatures at these pressure limits.

Higher temperature corresponding to pressure 1.8 M pa

$T_{H} = 62.9$ °c = 335.9 K

Lower temperature corresponding to pressure 0.2 M pa

$T_{L} = - 10.1$ °c = 262.9 K

COP of the system is given by

$COP = \frac{T_{L} }{T_{H} -T_{L} }$

$COP = \frac{335.9}{335.9 -262.9}$

COP = 4.6

Therefore the COP of the system is = 4.6

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3 years ago

Un acróbata de 60.0 kg está unido a un cordón de bungee con un resorte de 10.0 m de longitud . Salta de un puente que abarca un

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Answer:

What kind of languege is this please?

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2 years ago

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

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3 years ago

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