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A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no

ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, $\vec{F_{net}} = \vec{F'} - \vec{F}$

$\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N$

Thus the force will be along the direction of force whose magnitude is higher

Therefore,

$\vec{F_{net}} = 7 N$ towards North

4 0

3 years ago

Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i

Naddik [55]

Answer:

r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

F = ma

G m M / r² = m a

The centripetal acceleration is given by

a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

v = d / t

The distance traveled Esla orbits, in a circle the distance is

d = 2 π r

Time in time to complete the orbit, called period

v = 2π r / T

Let's replace

G m M / r² = m a

G M / r² = (2π r / T)² / r

G M / r² = 4π² r / T²

G M T² = 4π² r3

r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

T = 1.03 10⁸ s

Let's calculate

r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

r = ∛ (21.44 10³⁵ / 39.478)

r = ∛(0.0543087 10 36)

r = 0.3787 10¹² m

r = 3.787 10¹¹ m

7 0

3 years ago

Q 1. Choose the correct answer.

Natalija [7]

Answer:

Dietz

Explanation:

He is the guy you must justt be smart and know stuff.

7 0

2 years ago

Find the velocity of a baseball thrown 78 m from third base to first base in 30 sec

ankoles [38]

Answer:

V=2.6m/s

Explanation:

velocity=Distance/time

V=78/30=2.6m/s

3 0

2 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0

3 years ago

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