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3 years ago

A 63 kg ice skater finishes her performance and crossed the finish line with a speed of 10.8 m/s.

Physics

1 answer:

Mars2501 [29]3 years ago

4 0

Answer:

10.1 m/s

Explanation:

Every moving body or an object has a momentum. The simplest way to calculate it is to multiply the mass of an object with its velocity (p = m•v). So, just after this ice skater finished her performance, her momentum was 63 kg•10.8 m/s which equals to 680.4 kg•m/s. Then, she was given a huge bouquet which changed her total mass and speed. The thing we need to know is that the momentum before receiving the bouqet and after that are the same. With the bouqet, the mass increased and now is 63 kg + 4.4 kg = 67.4 kg. If p=m•v, then v=p/m, so her new speed is 680.4/67.4 = 10.0949, or approximately 10.1 m/s.

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A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

3 years ago

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F

yarga [219]

Answer:

Explanation:

mass, m = 12 kg

Force, F = 40 N

θ = 37° below the horizontal

(a)

Diagram is attached

(b) Let N be the normal reaction

According to the diagram

N + F Sin θ = m g

N = mg - F Sin θ

N = 12 x 9.8 - 40 x Sin 37

N = 117.6 - 24.07

N = 93.53

6 0

3 years ago

What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca

Katyanochek1 [597]

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{42}\textrm{K}$ = 41.962403 u

Mass of $_{20}^{42}\textrm{Ca}$ = 41.958618 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(41.962403-41.958618)=0.003785u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.003785u)\times c^2$

$E=(0.003785u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=3.526MeV$

Hence, the energy released in the given nuclear reaction is 3.526 MeV.

7 0

4 years ago

(2)science who ever gets this right will get a brainlest

const2013 [10]

Answer:

True

Explanation:

Cigarettes release over 5000 different chemicals when they burn and at least 70 of these can cause cancer.

Hope this helped!!!

7 0

3 years ago

Read 2 more answers

What is the defining feature of isotopes of the same element?

pogonyaev

Answer: Isotopes. An isotope is one of two or more forms of the same chemical element. Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number, but a different number of neutrons giving each elemental isotope a different atomic weight.

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6 0

2 years ago