1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Valentin [98]
4 years ago
5

Ignoring air resistance. during a projectiles motion the horizontal velocity will _____ as it travels.

Physics
1 answer:
Wittaler [7]4 years ago
5 0
The horizontal velocity will decrease as it travels.

Hope this helps you!
You might be interested in
Two of the types of infrared light, ir-c and ir-a, are both components of sunlight. their wavelengths range from 3000 to 1,000,0
NikAS [45]
The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
<span>6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
</span>6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J

For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J
6 0
3 years ago
Read 2 more answers
What method has detected the most extrasolar planets so far?
Sedaia [141]
<h2>Answer: The Transit  method</h2>

Detecting extrasolar planets by direct observation (with a telescope) is a complicated task. This is because any planet constitutes an extremely dim light source compared to the star around which it orbits.

So, to detect this extremely dim source is quite difficult due to the glare of the star's light that dulls it.

In this sense, scientists and astronomers have made several methods to find these extrasolar planets, among which the most successful has been the transit method.

This method is based on <u>astronomical transit</u>, a phenomenon in which a body (a planet in this case) passes in front of a larger one (the star), blocking (eclipsing) its vision to some extent.

It should be noted that this is the method currently used in the search for extrasolar planets. Space agencies such as ESA (Europe) and NASA (USA) have put into orbit satellites with extremely sensitive photometric sensors to observe even the smallest variations of intensity of a star due to the passage of a planet.

5 0
3 years ago
Raising the temperature of a gas will increase its pressure to<br> volume of the gas would do what
joja [24]

Answer: Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.

Explanation: In the 1600s, Boyle measured the volumes of gases at different pressures. Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.

3 0
2 years ago
NEWTONS SECOND LAW LAB REPORT
hram777 [196]

Answer:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Explanation:

here this may help.

8 0
3 years ago
A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio
goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

3 0
3 years ago
Other questions:
  • A solid copper cube is attached to spring-like device and is able to oscillate horizontally with negligible friction. Each side
    10·1 answer
  • A straight wire lies along the y-axis initially carrying a current of 10 A in the positive y-direction. The current decreases an
    6·1 answer
  • If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues tr
    12·2 answers
  • I need help on 2/3 please asap thx ♥️
    13·1 answer
  • What is the primary energy source that drives earth’s water cycle between the atmosphere, oceans and land?
    12·1 answer
  • HELP PLEASE!!!!! pulleys, wheels, and axels.
    15·2 answers
  • A mass of 100 grams of a particular radioactive substance decays according to the function m(t)=100e−t850
    13·1 answer
  • I will give 50 pts and brainliest
    8·1 answer
  • How are mass, distance and gpe related
    9·1 answer
  • Explain the relation between Joule and Erg.​
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!