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Troyanec [42]
2 years ago
10

. The gravitational force experienced by an object is. mass. weight.push .pull​

Physics
2 answers:
Bingel [31]2 years ago
8 0
Push
I think that’s the answer
Kitty [74]2 years ago
3 0
Bonjour,

the gravitational force experienced by an object is called the weight of the object.
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Water waves, earthquake waves, sound waves, and the waves that travel down a rope or spring are all examples of what waves.
Margaret [11]
Compression waves :-)
5 0
3 years ago
What is the momentum of a 3 kg bowling ball moving at 3 m/s?
Nataly [62]

Explanation:

<h3>p = mv</h3>

  • <em>p</em> denotes momentum
  • <em>m</em> denotes mass
  • <em>v</em> denotes velocity

→ p = 3 kg × 3 m/s

→ <u>p</u><u> </u><u>=</u><u> </u><u>9</u><u> </u><u>kg</u><u>.</u><u>m</u><u>/</u><u>s</u>

<u>Option</u><u> </u><u>D</u><u> </u><u>is</u><u> </u><u>corre</u><u>ct</u><u>.</u>

5 0
3 years ago
80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te
vodka [1.7K]

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

Therefore, the resulting temperature is 23.37 ⁰C

6 0
3 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
3 years ago
Which condition must exist in order for conduction to occur between two substances
motikmotik
Heat!!!!!!!!!!!!!!!!!!! sorry about the exclamation marks but it wont let me post in less it has more than 20 characteristics
6 0
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