Which statement best describes how light behaves with liquids, gases, and solids?

Which of the following is an example of exothermic reaction?

Katarina [22]

Explanation:

Exothermic reaction are those in which heat releases during a reaction

6 0

3 years ago

In the data table , distance is measured in meters and time is in seconds. Calculate the mans average velocity using the equatio

Artist 52 [7]

Answer:

3.626 m/s

Explanation:

v=d/t

1. -0.02/0 = 0 m/s

2. 0.86/0.2 = 4.3 m/s

3. 1.71/0.4 = 4.275 m/s

4. 2.54/0.6 = 4.23 m/s

5. 3.32/0.8 = 4.15 m/s

6. 4.08/1.0 = 4.08 m/s

7. 4.79/1.2 = 3.99 m/s

8. 5.48/1.4 = 3.91 m/s

9. 6.15/1.6 = 3.84 m/s

10. 6.76/1.8 = 3.76 m/s

11. 7.37/2.0 = 3.66 m/s

12. 7.92/2.2 = 3.6 m/s

13. 8.45/2.4 = 3.52 m/s

14. 8.96/2.6 = 3.45 m/s

the mean of these numbers is 3.626

his average velocity ks 3.626 m/s

6 0

3 years ago

Daffy Duck is standing 6.8 m away from Minnie Duck. The attractive gravitational force between them is 5.4x10-8 N. If Daffy Duck

artcher [175]

Answer:

432.78 Kg

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

8 0

3 years ago

Josie walks 15 meters north down the freshman hallway. She then walks 25 meters east. Lastly, she walks 15 meters south. What is

hichkok12 [17]

Answer:

the answer is yours a calculated

3 0

3 years ago

2

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0

2 years ago