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SVEN [57.7K]
3 years ago
10

What gravitational force does the earth exert on a person

Physics
1 answer:
lara [203]3 years ago
7 0

The  gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

<h3>What is the gravitational force of the earth on the person?</h3>

The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:

  • F = \frac{Gm^{1}m^{2}}{r^{2}}

where

G = 6.67 * 10⁻¹¹

m¹ = 62 kg

m² = 5.97 * 10²⁷ kg

r = 6.4 * 10⁶ m

F = \frac{5.97*10^{24}*62*6.67*10^{-11}}{(6.4*10^{6}){2}} = 602.74\:N

Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

Learn more about gravitational force at: brainly.com/question/940770

#SPJ1

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If Hubble's constant is 73 km/s/Mpc then the age of the universe is 13.4 billion years, assuming a uniform expansion. Suppose it
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Answer:

It would decrease the calculated age of the universe

Explanation:

Since the age of the universe is the reciprocal of Hubble's constant, it's therefore if Hubble's constant is increased the age decreases but if the Hubble's constant is decreased, the age of universe increases. Therefore, the age of universe and Hubble's constant are inversely proportional. Conclusively, any attempt to increase Hubble's constant would imply the calculated age of the universe decreases.

3 0
3 years ago
A 4.0 cm object is 6.0 cm from a convex mirror that has a focal length of 8.0 cm. What is the distance of the image from the mir
ioda

Mirror formula

1/f = (1/p) + (1/q)    [ f= focal length  ,  p= object distance ,  q = image distance ]

as given convex mirror so

distance of object =p= 6cm

focal length =  -8cm

to find q=?

1/q =( 1/f) - (1/ p)

1/q = (1/ 8) -(1/6)

1/q =-1/24

q=-24  ( negative sign shows image is virtual and behind the mirror)

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4 years ago
Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg
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Under the Big Top elephant, Ella (2500 kg), is attracted to Phant, the 3,000 kg elephant. They are separated by 8

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3 years ago
Insert
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

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