Từ độ cao h=2m một vật bắn lên với vận tốc ban đầu V0=10(m/s), hợp với phương ngang 1 góc 30 độ.Hãy xác minh

Snell's Law: Light traveling through water comes to a glass surface at an angle of incidence of

Lerok [7]

Answer:

1. The best definition of refraction is ____.

a. passing through a boundary

b. bouncing off a boundary

c. changing speed at a boundary

d. changing direction when crossing a boundary

Answer: D

Bouncing off a boundary (choice b) is reflection. Refraction involves passing through a boundary (choice a) and changing speed (choice c); however, a light ray can exhibit both of these behaviors without undergoing refraction (for instance, if it approaches the boundary along the normal). Refraction of light must involve a change in direction; the path must be altered at the boundary.

6 0

3 years ago

Describe all the ways that newtons laws can apply in a car crash

Harman [31]

Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash

Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.

Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her

7 0

3 years ago

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the

kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

$\theta=tan^{-1}(\frac{v_y}{v_x})$ (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

$v_x=v_o$

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

$v_y^2=v_{oy}^2+2gy$ (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

$v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}$

Finally, you use the equation (1) to find the angle:

$\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°$

The angle below the horizontal is 12.60°

7 0

3 years ago

Light wands are small tubes that can be used to give off light without electricity or fire.

miv72 [106K]

<span>c.the chemical energy of the fluids inside the wand</span>

4 0

3 years ago

Read 2 more answers

Learning Goal: To review the concept of conservative forces and to understand that electrostatic forces are, in fact, conservati

CaHeK987 [17]

Explanation:

The electrostatic forces are conservative forces!

The mainly property of the conservative fields is $\vec{\nabla} \times \vec E=\vec 0$

In spherical coordinates the field's expression is:

$\vec E=\frac{Q}{4\pi \epsilon _0 r^2} .\^r$

and the curl expression is:

$\nabla\times \vec E=\frac{1}{r^2{\sin}\,\theta}\left|\begin{matrix}\hat{r} & r\,\hat{\theta} & r\,{\sin}\,\theta\,\hat{\varphi} \\& & \\\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi}\\ & & \\E_r & rE_\theta & r{\sin}\,\theta\, E_\varphi\end{matrix}\right|=(0, 0, 0)$

to find the expression for the potential function associated:

$\vec E=\vec \nabla . V, \Delta V= V_b-V_a=-\int _c \vec E.d\vec l=-\int _c E\^r.dr\^r=-\int _c Edr=\int \limits^a_b \frac{Q}{4\pi \epsilon _0 r^2} dr= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{r}|^b_a)= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{b}-\frac{1}{a})$

5 0

3 years ago