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3 years ago

The acceleration due to gravity on Jupiter is 23.1 m/s2, which is about twice the acceleration due to gravity on Neptune.

Physics

2 answers:

Montano1993 [528]3 years ago

8 0

Answer:

C-An object weighs about two times as much on Jupiter as on Neptune.

Explanation:

on edg

polet [3.4K]3 years ago

3 0

Answer:

I believe its C

Explanation:

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A dog walks 26m, East, 29m, West, and 7m, East. What is his displacement?

sergeinik [125]

Answer:

4m East

Explanation:

26-29= -3 (so that means 3m west)

-3+7= 4 (so 4m East)

hope it helps

7 0

3 years ago

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha

trapecia [35]

Answer:

Velocity will be $v=4.06\times10^7m/sec$

Explanation:

We have given mass of alpha particle m = 4 u $=4\times 1.67\times 10^{-27}kg=6.68\times 10^{-27}kg$

Charge on alpha particle $q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C$

Charge on thorium particle $=90\times 1.6\times 10^{-19}=144\times 10^{-19}C$

Diameter is given as d = 15 fm

So radius $r=\frac{d}{2}=\frac{15}{2}=7.5fm=7.5\times 10^{-15}m$

Potential energy is given by $E=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r}=\frac{Kq_1q_2}{r}=\frac{9\times 10^9\times 144\times 10^{-19}\times 3.2\times 10^{-19}}{7.5\times 10^{-15}}=5.529\times 10^{-12}J$

From energy conservation $\frac{1}{2}mv^2=5.529\times 10^{-12}$

$\frac{1}{2}\times 6.68\times 10^{-27}v^2=5.529\times 10^{-12}$

$v=4.06\times10^7m/sec$

6 0

3 years ago

The production of heat by metabolic processes takes place throughout the volume of an animal, but loss of heat takes place only

Sever21 [200]

To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume

The initial measures,

$\text{Initial Length} = L$

$\text{Initial surface Area} = 6L^2$ (Surface of a Cube)

$\text{Initial Volume} = L^3$

The final measures

$\text{Final Length} = L_f$

$\text{Final surface area} = 6L_f^2$

$\text{Final Volume} = L_f^3$

Given,

$\frac{(SA)_f}{(SA)_i} = 2$

Now applying the same relation we have that

$(\frac{L_f}{L_i})^2 = 2$

$\frac{L_f}{L_i} = \sqrt{2}$

The relation with volume would be

$\frac{(Volume)_f}{(Volume)_i} = (\frac{L_f}{L_i})^3$

$\frac{(Volume)_f}{(Volume)_i} = (\sqrt{2})^3$

$\frac{(Volume)_f}{(Volume)_i} = (2\sqrt{2})$

$\frac{(Volume)_f}{(Volume)_i} = 2.83$

Volume of the cube change by a factor of 2.83

6 0

4 years ago

One boat tows another boat by means of a tow line, which is under a constant tension of 465 N. The boats move at a constant spee

Ludmilka [50]

Answer:

Work done, W = 141174 Joules

Explanation:

It is given that,

Constant tension acting on the boat, T = F = 465 N

Speed of the boat, v = 4.6 m/s

Time, t = 1.1 min = 66 seconds

Let W is the work done by the tension. It is equal to the product of force and displacement. It is given by :

$W=F\times d$

Since, $d=vt$

$W=F\times v\times t$

$W=465\ N\times 4.6\ m/s\times 66\ s$

W = 141174 Joules

So, the work is done by the tension is 141174 Joules. Hence, this is the required solution.

6 0

3 years ago

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

3 0

3 years ago