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ikadub [295]
2 years ago
7

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

angle of projection are kept the same. On Earth, the javelin covers a distance y. What distance would it cover on the moon? (The moon’s gravity is about 1/6th that of Earth.)
Physics
2 answers:
kodGreya [7K]2 years ago
7 0
x =  (v_{0}) ^{2} *   \frac{sin 2 \alpha}{g} + x_{0}

If g goes down by a factor of 6, the distance x goes up by a factor of 6.
mario62 [17]2 years ago
4 0

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

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A proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
lawyer [7]

The change in potential energy of the proton is  5.6 x 10^{-17} Joule

<h3>What is a Uniform Electric Field ?</h3>

A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.

Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

  • The distance d = 10 cm = 0.1 m
  • Electric field E = 3.5 KN/C
  • Proton charge q = 1.6 x 10^{-19} C

The Work done = Fd

but F = Eq

Recall that Electric field E = F/q = V/d

Where V = potential difference.

Let us first calculate the V

E = V/d

V = Ed

Substitute all the parameters into the formula above

V = 3.5 × 10³ × 0.1

V = 350 v

from F/q = V/d

make F the subject of formula and substitute it in work formula

F = Vq/d

W.D = Vq/d x d

W.D = Vq

Substitute all the parameters into the formula above

W.D = 350 x 1.6 x 10^{-19}

W.D = 5.6 x 10^{-17} J

Work done = Energy = Potential Energy

Therefore, the change in potential energy of the proton is 5.6 x 10^{-17}<em> Joule</em>

<em />

Learn more about Electric Field here: brainly.com/question/14372859

#SPJ1

7 0
2 years ago
A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0
algol13

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats m_1 and m_2,  and the length of the rod L.

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$,

solving for v, we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for m_1, m_2, g, and L we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}} $

\boxed{ v= 1.64m/s}

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

7 0
3 years ago
The earth travels around the sun once a year in an approximately circular orbit whose radius is 1.50x10^11 m. From this data det
seraphim [82]
(a) Determine the circumference of the Earth through the equation,
            C = 2πr
Substituting the known values, 
           C = 2π(1.50 x 10¹¹ m)
             C = 9.424 x 10¹¹ m

Then, divide the answer by time which is given to a year which is equal to 31536000 s. 
          orbital speed = (9.424 x 10¹¹ m)/31536000 s

               orbital speed = 29883.307 m/s

Hence, the orbital speed of the Earth is ~29883.307 m/s.

(b) The mass of the sun is ~1.9891 x 10³⁰ kg. 
8 0
3 years ago
What is the smallest unit that makes up matter
hram777 [196]
The smallest unit that makes up matter is an atom!
3 0
3 years ago
Read 2 more answers
A car accelerates from zero to a speed of 110
Verizon [17]

The car's rate of  acceleration : a = 2.04 m/s²

<h3>Further explanation</h3>

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

3 0
3 years ago
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