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2 years ago

What will be the gravitational force of two bodies when distance between them is doubled

Physics

1 answer:

Vladimir79 [104]2 years ago

8 0

When you double the distance the force becomes a quarter of what it was before.

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A bowling ball travels at 2.0 m/s. It has 16 J of kinetic energy. what Is the mass of the bowling ball in kilograms ?

zaharov [31]

Kinetic energy = 1/2 × m × v^2
16 = 1/2 × m × 2^2
16 = 1/2 × m × 4
16 = 2 × m
16/2 = m
8 = m

so the mass is 8 kg

5 0

3 years ago

Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnect

Iteru [2.4K]

Answer:

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

C = 1.0 + 1.0

C = 2.0 F

R = resistance = 33 Ω

Time constant is given as

T = RC

T = 33 x 2

T = 66 sec

V₀ = initial potential difference across the combination = 6.0 Volts

V = final potential difference = 2.2 volts

Using the equation

$V = V_{o} e^{\frac{-t}{T}}$

$2.2 = 6 e^{\frac{-t}{66}}$

t = 66.2 sec

5 0

3 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

vovikov84 [41]

Gravitational force is given by, $F= G\frac{mM}{R^{2}}$

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, $F_{1}= G\frac{m_{1}M}{R^{2}}$

Gravitational force of the star on planet 2, $F_{2}= G\frac{3m_{1}M}{(3R)^{2}}$

Ratio, $\frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}$

$\frac{F_{1}}{F_{2}}= \frac{3}{1}$

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0

2 years ago

Read 2 more answers

Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.

Gelneren [198K]

Answer:

$2.99\cdot 10^{11}C$

Explanation:

The mass of one electron is

$m_e = 9.11\cdot 10^{-31}kg$

So the number of electrons contained in M=1.7 kg of mass is

$N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}$

The charge of one electron is

$e=1.60\cdot 10^{-19} C$

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

$Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C$

8 0

2 years ago

The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of

PolarNik [594]

The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions

7 0

2 years ago

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What will be the gravitational force of two bodies when distance between them is doubled​

What will be the gravitational force of two bodies when distance between them is doubled