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lapo4ka [179]
2 years ago
9

What is projectile motion​

Physics
2 answers:
andrezito [222]2 years ago
6 0

Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path.

The path that the object follows is called its trajectory.

Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity

Radda [10]2 years ago
5 0

\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}

\sf{\qquad{\qquad{\underline{\underline{ Projectile~motion }}}}}

If an object is given an initial velocity in any direction and then allowed to travel freely under gravity only, it is called a projectile motion.

It is basically 3 types

  1. horizontally projectile motion
  2. oblique projectile motion
  3. included plane projectile motion

The path followed by a projectile is called its trajectory.

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HELP!!!
Vitek1552 [10]
We shall consider two properties:
1. Temperature difference
2. Thermal conductivity of the material

Use a cylindrical rod of a given material (say steel) which is insulated around its circumference.

One end of the rod is dipped in a large reservoir of water at 100 deg.C and the other end is dipped in water (with known volume) at 40 deg. C. The cold water if stored in a cylinder which is insulated on all sides. A thermometer reads the temperature of the cold water as a function of time.

This experiment will show that
(a) heat flows from a region of high temperature to a region of lower temperature.
(b) The thermal energy of a body increases when heat is added to it, and its temperature will rise.
(c) The thermal conductivity of water determines how quickly its temperature will rise. If mercury replaces water in the cold cylinder, its temperature will rise at a different rate because its thermal conductivity is different.



5 0
3 years ago
Read 2 more answers
A compass needle acts as a(n)
beks73 [17]

Explanation:

A compass needle acts as a Magnet

6 0
2 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

8 0
3 years ago
A block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane. (a) What is its velocity when it reaches t
kramer

Answer:

A.) 8 m/s

B.) 7.0 m

Explanation:

Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.

(a) What is its velocity when it reaches the top of the plane?

Since the plane is frictionless, the final velocity V will be the same as 8 m/s

The velocity will be 8 m/s as it reaches the top of the plane.

(b) How far horizontally does it land after it leaves the plane?

For frictionless plane,

a = gsinø

Acceleration a = 9.8sin28

Acceleration a = 4.6 m/s^2

Using the third equation of motion

V^2 = U^2 - 2as

Substitute the a and the U into the equation. Where V = 0

0 = 8^2 - 2 × 4.6 × S

9.2S = 64

S = 64/9.2

S = 6.956 m

S = 7.0 m

4 0
2 years ago
Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley
kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = \frac{4m}{2} x^2  + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2

T = 4mx^2 + 2my^2 -2mxy  

also the general potential energy can be expressed as

U = -4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant

The Lagrangian of the problem can now be setup as

L =4mx^2 +2my^2 -2mxy +2mgy + constant

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange  equation = 4x-y =0\\-2y+x +g = 0

solving the equations simultaneously

x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0
3 years ago
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