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3 years ago

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Calculate the force of gravity on the same 1 kg mass if it were 6.4x10^6 m above earth's surface

1 answer:

vova2212 [387]3 years ago

3 0

The force of gravity on a certain object is calculated through the equation,

F = Gm1m2 / r²

where F is the force, G is a constant, m1 and m2 are masses of the object and Earth, respectively and r is the distance. Substituting the known values for this item,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(1 kg)(5.98 x 10²⁴ kg)/ (6.4 x 10⁶ m)²

F = 9.37 N

Answer: 9.37 N

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Tanya [424]

D. 51 N. The minimum applied force that will cause the television slide is 51 N.

In order to solve this problem we have to use the force of static friction equation Fs = μs*n, where μs is the coefficient of static friction, and n is the normal force m*g.

With μs = 0.35, and n = 15kg*9.8m/s² = 147 N

Fs = (0.35)(147 N)

Fs = 51.45 N

Fs ≅ 51 N

3 0

3 years ago

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The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo

hjlf

Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ =p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

Finally we have:

eΔV=p^2/2m= h^2/(2*m*λ^2)

replacing we obtained the above values.

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3 years ago

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

Colt1911 [192]

Answer:

$0.53\times 10^{-30}kgms^{-1}$

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

$\Delta x\Delta p=\frac{h}{4\pi }$

Given that the uncertainty in x is 0.1 mm.

Therefore,

$\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}$

Therefore, uncertainty in the transverse momentum of photon is $0.53\times 10^{-30}kgms^{-1}$

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3 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ $ln[\frac{P_{1} }{P_{2} }]$

= <em>mRT</em>₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵<em>pV</em> = <em>mRT</em>]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = Δ<em>U </em>+ <em>W</em>

= 0 + 270.588

= 270.588 kJ

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3 years ago

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Aliun [14]

Answer:

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Explanation:

4 0

3 years ago

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