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Licemer1 [7]
3 years ago
7

What would the person’s mass be on the earth? Part B pls

Physics
1 answer:
Dafna1 [17]3 years ago
4 0

Answer:

I hopes it helps

thank you

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Alex pushes on a 2.0 kg book, resulting in a net force of 6.0 N on the book.
Yakvenalex [24]

Answer:

<h2>3.0 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a  = \frac{f}{m}  \\

From the question we have

a =  \frac{6}{2}  \\

We have the final answer as

<h3>3.0 m/s²</h3>

Hope this helps you

4 0
3 years ago
Mass is a fundamental physical quantity give scientific reason​
Lena [83]

Answer:

yes

Explanation:

because it is independent of eachother

8 0
2 years ago
The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 d
melisa1 [442]

This question is incomplete

Complete Question

The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 degrees Fahrenheit. Classify the objects according to whether their molecules will speed up or slow down after being left in the kitchen for a period of time.

Drag the objects to the correct category.

Object Temperature (Fahrenheit)

ice cubes 26°F

glass of tea 60°F

cooked piece of meat 160°F

butter 55°F

pot of water 75°F

bowl of soup 140°F

Answer:

Molecules that would speed up

ice cubes 26°F

glass of tea 60°F

butter 55°F

Molecules that would slow down

cooked piece of meat 160°F

pot of water 75°F

bowl of soup 140°F

Explanation:

Molecules that are found or contained in a substances have the tendency to react in such as what that they are are sped up or slowed down. This is due to their exposure to various kinds of changes in temperature, which could be a hot temperature , cold temperature e.t.c

When a substance is exposed to a hot temperature, the molecules of that substance speed up depending on how hot the temperature is while when a substance is cooled down or exposed to cold temperature, the molecules tend to slow down.

In the question above, the temperature of the kitchen is 72 degrees Fahrenheit.

Molecules that would speed up after been left in the Kitchen for a while:

a) Ice cubes 26°F: This is because 72°F is a warm took temperature, so the ice cubes would melt causing the molecules to speed up.

b)Glass of tea 60°F: This glass of tea is at a cool temperature of 60°F, when it is kept in the Kitchen which has a temperature of 72°F for while, the temperature of the glass of tea would increase due to an increase in its rate of reaction causing the molecules of the glass of water to speed up.

c) Butter 55°F: This is because 72°F is a warm temperature, so the butter would melt, increasing its rate of reaction and causing the molecules to speed up

Molecules that would slow down

a) Cooked piece of meat 160°F : 160°F is a very hot temperature and when it is left in a the Kitchen(72°F) , the cooked meat would begin to cool down and drop in temperature from 160°F, causing the molecules to slow down.

b) Pot of water 75°F: The temperature of the kitchen and the temperature of the pot of water is the same, hence the molecules of the water would slow down.

c) Bowl of soup 140°F : This bowl of soup is very hot and when it is left in the Kitchen(72°F) , the bowl of soup would begin to cool down and drop in temperature from 140°F, slowing down the rate of reaction and causing the molecules to slow down.

3 0
3 years ago
A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top
Advocard [28]

Answer:

a)  Em₀ = 42.96 104 J , b)   W_{fr} = -2.49 105 J , c)  vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

        Em = K + U

a) Let's look for the initial mechanical energy

      Em₀ = K + U

      Em₀ = ½ m v2 + mg and

      Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

      Em₀ = 36 104 + 6.96 104

      Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

    W_{fr} = Em₂ -Em₀

     Em₂ = K + U

     Em₂ = ½ m v₂² + m g y₂

     Em₂ = ½ 50 85 2 + 50 9.8 427

     Em₂ = 180.625 + 2.09 105

     Em₂ = 1,806 105 J

     W_{fr} = Em₂ -Em₀

     W_{fr} = 1,806 105 - 4,296 105

     W_{fr} = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

       W_{fr} = ΔEm

       W_{fr} = Emf - Emo

       -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

       ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

       ½ 50.0 vf² = 0.561

       vf = √ 0.561 25

      vf = 3.75 m / s

6 0
3 years ago
onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re
solmaris [256]

Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA =  1.00 * 10⁻⁴ * 1

Number of moles of HA =  1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺  +  A⁻

If 1 mol of HA produces 1 mol of H⁺  and  A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of  H⁺  and  A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of  H⁺ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of  A⁻ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into  H⁺  and  A⁻, 84% of HA is left

Number of mols of HA left = 0.84 *  1.00 * 10⁻⁴

Number of mols of HA left =  8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ =  1.6 * 10⁻⁵/V

To calculate the dissociation constant

k_{a} = [H^{+} ][A^{-} ]/[HA]

k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L

3 0
3 years ago
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