PLEASE I NEED THIS NOW ITS EASY I JUST DIDNT STUDY

An electromagnetic wave is produced by:

Ray Of Light [21]

A. an accelerating charged charged particle or changing magnetic fields

4 0

2 years ago

What happens to light when it travels from air into water

spin [16.1K]

Answer:

Water is more dense than air. When water goes through a denser thing, the light is "bent" more towards the "normal" which is a straight, vertical line.

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8 0

2 years ago

A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4

Vedmedyk [2.9K]

Hello!

Vx = V0x + Ax*t
Vx = 18.1 + 2.4t

Let’s take time as 7.50 seconds:
Vx = 18.1 + 2.4*7.50
Vx = 18.1 + 18 = 36.1 m/s

Then, the final velocity of the car is 36.1 m/s.

3 0

3 years ago

A natural force of attraction exerted by the earth upon objects, that pulls

Mashutka [201]

A natural force of attraction exerted by the earth upon objects, that pulls objects towards earth's center is called Gravitational force .

6 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

2 years ago