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Fantom [35]
3 years ago
13

What is the gravitational force between a 45 kg person, and the Earth at 5.98 x 1024 kg, with a distance of

Physics
1 answer:
Assoli18 [71]3 years ago
7 0

Answer:

C. 441 N

Explanation:

Gravitational force between two objects can by calculated by the formula

= G m₁m₂ / r² , m₁ and m₂ are masses at distance r

= ( 6.67 x 10⁻¹¹ x 45 x 5.98 x 10²⁴) / ( 6.38 x 10⁶ )²

= 44.09 x 10

= 440.9 N

= 441 N .  

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A 60 kg sprinter has a momentum of +600 kg-m/s when he crosses the finish
MakcuM [25]

Answer:

10 ms⁻¹

Explanation:

The amount of momentum that an object has is dependent upon two factors

  • mass of the moving object  
  • speed of motion

In terms of an equation,

Momentum (P) = Mass(m)×velocity(v)

                     P = m×v

                 600 = 60 × v ⇒ v = 10 ms⁻¹

3 0
3 years ago
A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?
posledela

Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

\displaystyle f = \frac{v}{\lambda},

where

  • v represents the speed of this wave, and
  • \lambda represents the wavelength of this wave.

6 0
3 years ago
Suppose there are 100,000 atoms of a radioactive substance that has a ½ life of 10 minutes. How many atoms will remain after 40
dezoksy [38]

Answer:

c. 12,500

Explanation:

Original number of atoms = 100,000 atoms

Half- life  = 10min

Unknown:

The number of atoms that will remain after 10min  = ?

Solution:

The half - life is the time taken for half of a radioactive substance to decay by half.

    Time taken        Number of atom   half life

           10min             100000                   _

           20min             50000                    1

           30min             25000                    2

           40min              12500                     3

6 0
3 years ago
Squids propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity
Musya8 [376]

Answer:

v_squid = - 2,286 m / s

Explanation:

This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.

Initial moment. Before expelling the water

          p₀ = 0

the squid is at rest

Final moment. After expelling the water

         p_{f} = M V_squid + m v_water

         p₀ = p_{f}

          0 = M V_squid + m v_water

           c_squid = -m v_water / M

The mass of the squid without water is

            M = 9 -2 = 7 kg

let's calculate

           v_squid = 2 8/7

           v_squid = - 2,286 m / s

The negative sign indicates that the squid is moving in the opposite direction of the water

8 0
4 years ago
Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to up
Burka [1]

Answer:

True

Explanation:

If there's no preference over the string case (upper case or lower case), one can convert both strings to upper case or to lowercase and then compare the converted strings to test if they're equal or not.

An Illustration is

string a = "Boy"

string b = 'bOy"

if(a.ToUpper() == b.ToUpper() || a.ToLower() == b.ToLower())

{

Print "Equal Strings"

}

else

{

Print "Strings are not equal";

}

The above will first convert both strings and then compare.

Since they are the same (after conversion), the statement "Equal Strings" will be printed, without the quotes

5 0
4 years ago
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