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3 years ago

When the Moon is in the position shown, how would the Moon look from Earth?

1 answer:

8 0

Answer:When the Moon is in the position shown below, how would the Moon look to an observer on the North Pole? The moon's rotational period is equal to its revolutional period around Earth. This statement is false because at night we see one side and during the day we see the opposite.

Explanation:

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Natural gas is very abundant in the United States. Compared to other fossil fuels, which describes the waste products of the com

SVETLANKA909090 [29]

Option d
. as they produce some hydrocarbon and methane etc

6 0

3 years ago

Read 2 more answers

How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y

vredina [299]

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity , $dN/dt = N* \lambda$

$\frac{dN}{dt} = N* \frac{ln2}{T_1/2}$

$N = \frac{(dN/dt )(T1/2)}{ln2}$

= ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

= 8.877*10^{16}

Number of moles:

n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

-----------------------------------------------------------------------------------------

time t = -[T1/2 / ln2]*ln[N/N0]

= - [5.3 years / ln2]*ln[1x10-6/1x10-3]

= 52.8 year

8 0

4 years ago

What amount of force is needed to propel and object of 27 kg to an acceleration of 11,550 m/s^2? (1 point)

aliya0001 [1]

Answer:

311,850 N

Explanation:

We can solve the problem by using Newton's second law:

$F=ma$

where

F is the net force applied on an object

m is the mass of the object

a is its acceleration

For the object in this problem,

m = 27 kg

$a=11550 m/s^2$

Substituting, we find the force required:

$F=(27)(11550)=311,850 N$

3 0

3 years ago

A piece of putty and a tennis ball with the same mass are thrown against a wall with the same velocity. Which object experiences

cluponka [151]

Answer:

Explanation:

Firstly, we have to define momentum.

Momentum is define as the product of mass and velocity.

That is P = mass×velocity

Also considering the third law of motion which states that: For every action, there is equal and opposite reaction.

Moreso, considering the 2nd law of motion which states that the rate of change in the momentum of a body is equal to the applied force and takes place in the direction of the applied force.

Now, applying P = mass×velocity

They both have same mass and velocity definitely, they will both experience same momentum.

Also from the question, the both share same velocity hence, the will both hit the wall with same velocity meaning the will both feel the same impact from the wall as well. Hence the third law of motion proves this right.

5 0

3 years ago

At the top of a giant swing on the gymnastics high bar, candy's velocity is 1 m/s, and she is 3.5 m high. if candy's mass is 50

Allushta [10]

The total mechanical energy is the sum of the kinetic energy and the gravitational potential energy:
$E=K+U= \frac{1}{2}mv^2 +mgh$
where m=3.5 kg is Candy's mass, v=1 m/s is her velocity and h=3.5 m is her height. If we replace these numbers, we find the mechanical energy of the system:
$E= \frac{1}{2} (50 kg)(1m/s)^2 + (50 kg)(9.81 m/s^2)(3.5 m)=1742 J =1.74 kJ$

6 0

3 years ago