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muminat
3 years ago
8

Some believe that the positions of the planets at the time

Physics
1 answer:
laila [671]3 years ago
7 0

Answer:

1.4007\times 10^{-8}\ N

1.50075\times 10^{-6}\ N

0.000000667\ N

Explanation:

m_1 = Mass of baby = 3 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Distance between objects

Gravitational force of attraction is given by

F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 70\times 3}{1^2}\\\Rightarrow F=1.4007\times 10^{-8}\ N

The force between baby and obstetrician is 1.4007\times 10^{-8}\ N

F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(6\times 10^{11})^2}\\\Rightarrow F=1.50075\times 10^{-6}\ N

The force between the baby and Jupiter is 1.50075\times 10^{-6}\ N

F=\dfrac{6.67\times 10^{-11}\times 3\times 2.7\times 10^{27}}{(9\times 10^{11})^2}\\\Rightarrow F=0.000000667\ N

The force between the baby and Jupiter is 0.000000667\ N

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3 years ago
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1. What happens to an atom when it gains electrons? *
Deffense [45]

Answer: An atom that gains or loses an electron becomes an ion. If it gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion

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3 years ago
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A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

3 0
3 years ago
A 7-kg bowling ball moving at 4 m/s strikes a 1 kg bowling pin. If the ball slows to 2 m/s in 0.05 s, how much force does it exe
iren [92.7K]

Answer:

280 N

Explanation:

acceleration = v2-v1 / time taken = (2-4 )/ 0.05 = -40 m/s^2  ( neg sign indicates slowing down )

force exerted = ma = 7 kg x -40 m/s^2 = - 280 N ( neg sign means opposite direction of initial velocity )

since the 7 kg ball is slowing down, the  direction of force will be opposite of the initial velocity , and it will be 280 N

7 0
3 years ago
What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
3 years ago
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