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
Fed [463]
3 years ago
14

Pls help my with this projectile motion project if you answer right ill give 75 points !!!!!!!!

Physics
2 answers:
Nata [24]3 years ago
7 0

Answer:

what is

Explanation:

huh I'm confused

WITCHER [35]3 years ago
3 0

Answer:

pumpkins 2 and 3 will hit the target. i did the math

Explanation:

You might be interested in
At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear
Mandarinka [93]
We apply the gravity calculation expressed in the formula: g=GM/r2 
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1)       Radius = √6.674e-11*5.972e24/8             = 7058 kms    Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2     Radius=√6.674e-11*5.972e24/4.9                 = 9019 kms        Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
4 0
3 years ago
1000 kg has 50, 000 joules of kinetic energy. what is it's speed?
kicyunya [14]
The answer is 36 kilometers per hour, or 10 meters per second.
7 0
3 years ago
The length of a 100 mm bar of metal increases by 0.3 mm when subjected to a temperature rise of 100°C. The coefficient of linear
Juli2301 [7.4K]

Answer:

α = 3×10^-5 K^-1

Explanation:

let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.

then, the coefficient of linear expansion is given by:

α = ΔL/(ΔT×L)

   = (0.3×10^-3)/(100)(100×10^-3)

   = 3×10^-5 K^-1

Therefore, the coefficient of linear expansion is 3×10^-5 K^-1

5 0
4 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
A heavy bank-vault door is opened by the application of a force of 300 N directed perpendicular to the plane of the door at a di
kondaur [170]
The answer is 240 Nm.
5 0
3 years ago
Other questions:
  • A lever is used to lift a heavy weight. The length of the input arm of the lever is 3 m, the length of output arm is 1 m.
    15·1 answer
  • When you double the distance between a pair of charged particles what happens to the force between them?
    5·1 answer
  • The colors of pigments is it a) are cyan,yellow,and magenta b) are the same as the secondary c) combine in equal amounts to prov
    12·1 answer
  • The mass of a sulfur atom is 32.0 u, and the mass of a fluorine atom is 19.0 u. What is the mass of a mole of sulfur hexafluorid
    8·1 answer
  • When two oppositely charged electrodes are immersed in a solution, positively charged ions are attracted to the negative electro
    12·1 answer
  • HELP FAST MARKING BRAINLEST A neutral electroscope is touched with a positively charged rod. After the rod is removed the electr
    7·1 answer
  • If the current density in a wire or radius R is given by J-k+5,0F wire? R, what is the current in the wire?
    7·1 answer
  • Three of your friends are all sitting g on one end of a seesaw. The combined weight is 275 N. The length from the fulcrum to you
    12·1 answer
  • Two rams run toward each other. One ram has a mass of 49 kg and runs west
    7·1 answer
  • Imagine that person B is more massive than person A in the picture above.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!