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3 years ago

Calculate the force between two objects that have masses of 70 Kilograms and 2,000 kilograms separated by a distance of 1 meter

Physics

1 answer:

Hunter-Best [27]3 years ago

5 0

9.3 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1 = 70kg

Mass 2 = 2000kg

distance = 1m

Unknown:

force between them =

Solution:

The force between the two masses will be a gravitational force of attraction.

F = $\frac{G m_{1}m_{2} }{r^{2} }$

G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²

r is the distance between the two masses

Substituting the parameters:

F = $\frac{6.67 x 10^{-11} x 70 x 2000}{1^{2} }$ = 9.3 x 10⁻⁶N

Learn more:

Universal gravitation constant brainly.com/question/1724648

#learnwithBrainly

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Answer:

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Explanation:

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3 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

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3 years ago

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N

irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

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Answer:

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