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3 years ago

13

The gravitational pull of Jupiter is greater than the gravitational pull of Earth. How would adding a “JUPITER MODE” to the virt

ual activity change the results of Galileo’s experiment?

1 answer:

Vlad1618 [11]3 years ago

5 0

The Galileo's experiment is about Earth’s gravitational force have on objects of different masses.

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A swimmer moves through the water at 2.5 meters per second and experiences a drag force of 240 N. How much power is she generati

mote1985 [20]

-- There's a force of 240N pushing her backwards.

-- She's maintaining a steady speed (of 2.5 m/s) .

-- In order to maintain a steady speed (no acceleration),
the forces on her must be balanced. So she's maintaining
a steady force of 240N forward.

-- Every time she moves 1 m forward, she does work of
(force) x (distance) = 240 joules.

-- She moves 2.5 meters forward every second.
So she's doing (240 x 2.5) = 600 joules of work every second.

-- 600 joules per second = 600 watts .

6 0

3 years ago

Hi how are u,here u go i will give brainliest!!!

elena55 [62]

Answer:

I'm good things for asking.

Explanation:

Hope you're having a great day too, and thank you for brainliest.

3 0

3 years ago

2 large parallel plates are 2.0 cm apart. the strength of the electric plates between them is 100.0 n/c . what is the difference

Citrus2011 [14]

Hello!

We can use the following equation for the potential difference between two parallel plates given an electric field and separation distance:
$V = Ed$

V = Potential Difference (? V)
E = Electric field strength (100.0 N/C)

d = distance between places (0.02 m)

Plug in the values and solve.

$V = (100)(0.02) = \boxed{2 V}$

4 0

2 years ago

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

B)

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

C)

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

3 years ago

Problem 9.22. what is the value of r (in ohms required for the circuit to be critically damped?

Goshia [24]

Critical damping is where there is no oscillations or overshoot but the thing is not overdamped. there's a quadratic in the equation somewhere which may help find r.

4 0

3 years ago