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notsponge [240]
3 years ago
10

True or false ??

Physics
2 answers:
galben [10]3 years ago
6 0
Well, gravity's acceleration is -9.81 m/s^2 so false.
Vinvika [58]3 years ago
3 0
False, because objects have different masses. Remember, mass determines acceleration
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Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains
lakkis [162]

Given that the density of heptane is

d_h=\frac{0.684g}{mL}

The mass of heptane is

m_h=31\text{ g}

The density of water is

d_w=\frac{1g}{mL}

The mass of water is

m_w=37\text{ g}

The volume of heptane will be

\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}

The volume of water will be

\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}

Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.

The total volume of liquid in the cylinder will be

\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}

The total volume of liquid in the cylinder will be 82.32 mL.

7 0
10 months ago
A certain computer chip that has dimensions of 3.67 cm and 2.93 cm contains 3.5 million transistors. If the transistors are squa
Evgesh-ka [11]

Answer:

1.56 × 10^-3 cm.

Explanation:

So, we are given the following parameters from the question above;

Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.

Step one: find the area of the computer chip.

Therefore, Area = Length × breadth.

Area = 3.67 cm × 2.93 cm.

Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.

Step two: find the area of one transistor

The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).

Hence;

The area of one transistor= 10.7531/4.4 × 10^6.

The area of one transistor= 2.44 × 10^-6 cm^2.

=> Note that We have our transistors as square, therefore;

The maximum dimension = √ (2.44 × 10^-6) cm^2.

The maximum dimension= 1.56 × 10^-3 cm.

7 0
3 years ago
Does air resistance affect the motion of a falling object differently when the initial velocity of the object is greater?
DIA [1.3K]
Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity. 
4 0
3 years ago
Read 2 more answers
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
2 years ago
16. Compared to your weight and mass on Earth, if you were on the moon: *
Natalija [7]

Answer:

The answer is D.

Explanation:

There is no gravity in Space so that means that it will decrease your weight but not your mass.

<h2><u><em>Please give Brainiest</em></u></h2>
5 0
3 years ago
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