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2 years ago

Which of the following exerts the strongest gravitational force on Earth? (Need help)

Physics

1 answer:

jarptica [38.1K]2 years ago

6 0

Answer:

Moon

Explanation:

although the moon is by far the smallest mass of the listed bodies, it is also by far the closest.

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How do greenhouse gases such as CO2 and N2O contribute to an increase in Earth’s atmospheric temperature?

konstantin123 [22]

CO2 and N2O keep the energy that gets to Earth from the sun inside the atmosphere. Without greenhouse gases, our planet would be too cold. But due to the recent increase in greenhouse gases, more energy released from the sun is contained in the Earth, heating it up.

3 0

3 years ago

Read 2 more answers

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

2 years ago

The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu

mel-nik [20]

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

$v=\sqrt{2.5\ r}$ .........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

$v=\sqrt{2.5\ \times 380}$

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

4 0

2 years ago

What does density have to do with heat?

joja [24]

If you take a fluid (i.e. air or water) and heat it, the portion that is heated usually expands. The same mass takes up more volume and as a consequence the heated portion becomes less dense than the portion that is<span><span> not heated.</span> </span>

8 0

2 years ago

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The 8.00-cm long second hand on a watch rotates smoothly.

dolphi86 [110]

The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s

4 0

2 years ago