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3 years ago

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Для вимірювання температури води, що має масу 66 г, у неї занурили термометр, який показав температуру 32,4 °С. Якою була дійсна

температура води, якщо теплоємність термометра 1,9 Дж\кг•°с і перед зануренням у воду він показував температуру навколишнього середовища рівну 17,8°С? Питома теплоємність води 4200 Дж\кг•°С

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

Answer:

я не знаю відповіді

Explanation:

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The nutritional label above came from a can of soup how many servings of soup would you need to eat to receive your recommended

Flauer [41]

Each serving gives 3 grams or 12% of daily recommended value. Divide 100% by 12% to get 8.33 servings to obtain the daily recommended value of 25 grams.

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3 years ago

What is the only possible value of ml for an electron in an s orbital?

Archy [21]

Answer:

zero

Explanation:

$m_l$ is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

$l$ : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

This quantum number is related to the type (or shape) of the orbital:

For s orbitals $l=0$

For p orbitals $l=1$

For d orbitals $l=2$

For f orbitals $l=3$

In this case, it is an s orbital, so we have $l=0$ .

$m_l$ , the third quantum number can have integer values ${from-l}$ to ${+l}$

Since, for the s orbitals $l=0$ , the only possible value for ${m_l}$ is zero.

4 0

3 years ago

A pogo stick has a spring with a force constant of 2.50 × 104 N/m , which can be compressed 11.2 cm. what maximum height, in met

AleksandrR [38]

Answer:

h = 36.4 cm

Explanation:

given,

spring constant = 2.5 x 10⁴ N/m

compressed distance = 11.2 cm = 0.112 m

mass of the child = 44 kg

maximum height = ?

by energy of conservation

$K.E_i +P.E_i= K.E_f + P.E_f$

$\dfrac{1}{2}kx^2 = mgh$

$\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h$

$\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h$

$156.8 = 44 \times 9.8 \times h$

$h = \dfrac{156.8}{44 \times 9.8}$

h = 0.364 m

h = 36.4 cm

7 0

3 years ago

a roller coaster begins at the top of a hill if it acelerates at the rate of 2 m/s2 and has a mass of 2000 kg what net force is

AlladinOne [14]

Answer:

$\boxed {\boxed {\sf 4000 \ Newtons }}$

Explanation:

Force can be found by multiplying the mass by the acceleration.

$F=m*a$

The mass of the roller coaster is 2000 kilograms and the acceleration is 2 meters per second squared.

$m= 2000 \ kg \\a= 2 \ m/s^2$

Substitute the values into the formula.

$F= 2000 \ kg * 2 \ m/s^2$

Multiply.

$F= 4000 \ kg*m/s^2$

1 kg*m/s² is equal to 1 N
Therefore our answer of 4000 kg*m/s² is equal to 4000 Newtons

$F= 4000 \ N$

The net force acting on the roller coaster is <u>4000 Newtons.</u>

7 0

3 years ago

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

7 0

2 years ago