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valentinak56 [21]
2 years ago
8

If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?  (

___/ 2)
Physics
1 answer:
valkas [14]2 years ago
8 0
The gravitational force between two masses m₁ and m₂ is
F=G \frac{m_{1} m_{2}}{d^{2}}
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d =  distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
     = 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm
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We are given a mercury atom in the ground state which absorbs 20 eV of energy. It is then ionized by losing an electron. We need to calculate the kinetic energy that the electron has after ionization.

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The electron charge is = 1.60217662 × 10-19<span> coulombs

To determine the kinetic energy, we can use this equation:

KE = 20 Joules / Coulombs * </span>1.60217662 × 10-19<span> coulombs 
KE = 1.25x10^20 Joules 

Therefore, the amount of kinetic energy that the electron has after ionization is </span>1.25x10^20 Joules or 1.25x10^17 kJ. <span />
8 0
2 years ago
Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?
MakcuM [25]
The angle of refraction would be further less 
3 0
2 years ago
A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an
inna [77]

Answer:

It will take 4 sec rock to comes its original point

Explanation:

It is given that the rock comes to its original point

So displacement S = 0 m

Initial velocity u = 19.6 m/sec

Acceleration due to gravity g=9.8m/sec^2

According to second equation of motion h=ut+\frac{1}{2}gt^2

0=19.6\times t+\frac{1}{2}\times 9.8t^2

19.6=4.9t

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3 0
2 years ago
Read 2 more answers
A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

5 0
2 years ago
The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin
KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50  , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t =  64 ,  ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms  .

8 0
2 years ago
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