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Pavlova-9 [17]
3 years ago
5

A 235 kg object and a 1.37×1012 kg are located 2.59×104 m away from each other. What is the force due to gravity between the two

objects?
Physics
2 answers:
V125BC [204]3 years ago
8 0

Answer:

the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Explanation:

Given;

mass of the first object, m₁ = 235 kg

mass of the second object, m₂ = 1.37 x 10¹² kg

distance between the two object, r = 2.59 x 10⁴ m

The gravitational force between the two object is calculated as;

F= \frac{Gm_1m_2}{r^2}

where;

G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²

F= \frac{(6.67\times 10^{-11})(235)(1.37\times 10^{12})}{(2.59\times 10^4)^2} \\\\F = 3.2 \times 10^{-5} \ N

Therefore, the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.

Alexus [3.1K]3 years ago
4 0

Answer:

F = 3.2 x 10⁻⁵ N

Explanation:

The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:

F = \frac{Gm_{1}m_{2}}{r^{2}}

where,

F = gravitational force = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m₁ = mass of object 1 = 235 kg

m₂ = mass of object 2 = 1.37 x 10¹² kg

r = distance between objects = 2.59 x 10⁴ m

Therefore,

F = \frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(235\ kg)(1.37\ x\ 10^{12}\ kg)}{(2.59\ x\ 10^{4}\ m)^{2}}

<u>F = 3.2 x 10⁻⁵ N</u>

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A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen
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Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

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The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s

N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns

Therefore, the number of turns of wire needed is 573.8 turns

4 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

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3 years ago
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Answer:

Explanation:

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p1 = 50(4) so

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Why does a lone pair of electrons occupy more space around a central atom than a bonding pair of electrons?
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Answer:

The lone pair of electrons occupy more space because the electrostatic force becomes weaker.

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When there is a bond pair of electrons in the 2 positively charged the atomic nuclei draw the electron density towards them, thereby reducing the bond diameter.

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