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liubo4ka [24]
3 years ago
11

The strength of an object's gravitational force is affected by both____ and____

Physics
1 answer:
DochEvi [55]3 years ago
3 0

Answer:

the answers are "mass" and "distance"

Explanation:

hope this helps, plz mark brainiest :)

You might be interested in
You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
serious [3.7K]

Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

8 0
3 years ago
Ice of mass 5 g at 0 °C melts to water at 0 °C.
amid [387]

Answer:

Q=1670J

Explanation:

Mass of ice: m=5g=0.005kg

Latent heat: lambda=3.34×10⁵J/kg

Heat received by ice: Q=m×lambda

Q=0.005×3.34×10⁵=5×334=1670J

5 0
3 years ago
Why dont we use a power water engine
devlian [24]
Cuz its slow for like to power the tv with water it will take about 2 days to turn on

7 0
3 years ago
Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c
PIT_PIT [208]
Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
E=k_e \frac{q}{r^2}
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C

While the electric field generated by the second charge is
E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C
4 0
3 years ago
a wave travels in a string at 58 m/s. a second string of 10% greater linear density has the same tension applied as in the first
ozzi

Answer:

The speed of wave in the second string is 55.3 m/s.

Explanation:

Given that,

Speed of wave in first string= 58 m/s

We need to calculate the wave speed

Using formula of speed for first string

v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}...(I)

For second string

v_{2}=\sqrt{\dfrac{T}{\mu_{2}}}...(II)

Divided equation (II) by equation (I)

\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\dfrac{T}{\mu_{2}}}{\dfrac{T}{\mu_{1}}}}

Here, Tension is same in both string

So,

\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{\mu_{2}}}

The linear density of the second string

\mu_{2}=\mu_{1}+\dfrac{10}{100}\mu_{1}

\mu_{2}=\dfrac{110}{100}\mu_{1}

\mu_{2}=1.1\mu_{1}

Now, Put the value of linear density of second string

\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{1.1\mu_{1}}}

v_{2}=v_{1}\times\sqrt{\dfrac{1}{1.1}}

v_{2}=58\times\sqrt{\dfrac{1}{1.1}}

v_{2}=55.3\ m/s

Hence, The speed of wave in the second string is 55.3 m/s.

5 0
4 years ago
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