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3 years ago

How is Uranus colder than Neptune

1 answer:

4 0

With Uranus at an average distance of 2.88 billion kilometres from the Sun and Neptune at an average distance of 4.5 billion kilometres it would be very easy to point out which of the gas giants is the coldest, but if you were you were to say that Neptune was the coldest, you’d be wrong.Given that we expect planets further from the Sun to be colder than those closer, this does make Neptune and Uranus quite a mysterious pair. Uranus and Neptune are brimming with volatiles such as water, methane and ammonia and due to their composition in comparison to Jupiter and Saturn, which are comprised mainly of hydrogen and helium, are labelled the ice giants. Scientists have measured how hot Uranus and Neptune should be and have found that Uranus is very cold and very dim

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What are the 5 steps in the process of air circulation in order

creativ13 [48]

Yes,Physics 5pts just noware the 5 steps in the process of air circulation in order

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3 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

olchik [2.2K]

Answer:

0.833 N

Explanation:

Formula for Kinetic Energy $E_k = \frac{mv^2}{2}$

Formula for Potential Energy $E_p = mgy$

First we need to find the vertical distance between the maximum-angle position and the pendulum lowest point:

Using the swinging point as the reference, the vertical distance from the maximum-angle (34 degree) position to the swinging point is:

$L * cos(34^o) = 1.2cos(34^o) = 1.2*0.83 = 0.995 \approx 1 m$

At the lowest position, pendulum is at string length to the swinging point, which is 1.2 m. Therefore, the vertical distance between the maximum-angle position and the pendulum lowest point would be

y = 1.2 - 1 = 0.2 m.

As the pendulum is traveling from the maximum-angle position to the lowest point position, its potential energy would be converted to the kinetic energy.

By law of energy conservation:

$E_k = E_p$

$\frac{mv^2}{2} = mgy$

$v^2 = 2gy$

$v = \sqrt{2gy}$

Substitute $g = 10 m/s^2$ and y = 0.2 m:

$v = \sqrt{2 * 10 * 0.2} = \sqrt{4} = 2 m/s$

At lowest point, pendulum would generate centripetal tension force on the string:

$F = m\frac{v^2}{L}$

We can substitute mass m = 0.25, rotation radius L = 1.2 m and v = 2 m/s:

$F = 0.25\frac{2^2}{1.2} = 0.833 N$

5 0

3 years ago

Question 25 of 25 What is a limitation of modeling the effect of gravity on a falling object by using a ball dropped from differ

agasfer [191]

Answer:

Air slows down The fall of any object including a ball.

5 0

3 years ago

Read 2 more answers

Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit

Reika [66]

The Answer is C Both at the same time

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3 years ago

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li

erica [24]

Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

Stan's speed = 42.9 m/s

Explanation:

Given:

$u_k$ = initial speed of Kathy = 0 m/s

$u_s$ = initial speed of Stan = 0 m/s
$a_k$ = acceleration of Kathy = $4.73\ m/s^2$

$a_s$ = acceleration of Stan = $3.9\ m/s^2$

Assumptions:

$v_k$ = final speed of Kathy when see catches Stan
$v_s$ = final speed of Stan when Kathy catches him
$s_k$ = distance traveled by Kathy to catch Stan
$s_s$ = distance traveled by Stan when Kathy catches him
$t_k$ = time taken by Kathy to catch Stan = $t$
$t_s$ = time interval in which Kathy catches Stan = $t+1$

Part (a):

Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

$\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\$

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

$s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5$

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

$v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3$

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

$v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9$

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

7 0

3 years ago