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SCORPION-xisa [38]
4 years ago
8

What science did the study of the night sky eventually become?

Physics
1 answer:
Andrews [41]4 years ago
7 0

This study of science is called Astronomy; the study of the night sky, planets, galaxies, and other celestial objects
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Practice 3: Label the correct phase that would result from the Moon and Earth in these positions.
Anna71 [15]

Answer:

both position I think in nor

5 0
3 years ago
Read 2 more answers
If the truck has a mass of 2,000 kilograms , what's its momentum?(v=35 m/s)
katen-ka-za [31]

Answer:

\boxed {\boxed {\sf  70,000 \ kg*m/s}}

Explanation:

Momentum is the product of mass and velocity.

p=m*v

The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.

m= 2000 \ kg \\v= 35 \ m/s

Substitute the values into the formula and multiply.

p= 2000 \ kg * 35 \ m/s \\p= 70,000 \ kg*m/s

The truck's momentum is <u>70,000 kilograms meters per second.</u>

8 0
3 years ago
A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘
hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of  mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution,T_f = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC

\Delta T_f=i\times K_f\times m

Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}

5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})

On solving we get:

\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of Ca(NO_3)_2 in the mixture:

\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%

The mass percentage of calcium nitrate is 31.23%.

5 0
3 years ago
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0
4 years ago
A 100-watt lightbulb uses 1 kilowatt-hour of electrical
Harlamova29_29 [7]

Answer:

Energy = 0.25 kilowatt-hour

Explanation:

Given the following data;

Power = 25 Watts

Time = 10 hours

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

To find the energy consumed;

Energy = power * time

Substituting into the formula, we have;

Energy = 25 * 10

Energy = 250 Watt-hour

To convert to kilowatt-hour, we would divide by 1000;

Energy = 250/1000

Energy = 0.25 kilowatt-hour

3 0
3 years ago
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