All categories

1 year ago

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

Physics

1 answer:

fgiga [73]1 year ago

8 0

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

#SPJ1

You might be interested in

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

sladkih [1.3K]

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0

2 years ago

The work done by the brakes during braking is equal to

kow [346]

Answer:

The amount of pressure delivered to them

Explanation:

6 0

1 year ago

Calculate the momentum of a 4 kg Bolling ball being thrown at a speed of 3 m/s

bonufazy [111]

We are given –

Mass of boiling ball is, m = 4 kg
Speed is, v = 3 m/s
Momentum, P =?

As we know –

↠Momentum = Mass × Speed(Velocity)

↠Momentum = 4 × 3 kgm/s

↠Momentum = 12 kgm/s

Henceforth,Momentum will be 12 kgm/s.

7 0

1 year ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

7 0

2 years ago

The greater the difference in electronegativity between two covalently bonded<br><br> atoms

katrin [286]

Answer:

The greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

Explanation:

Bond polarity (i.e the separation of electric charge along a bond) and ionic character (amount of electron sharing) increase with an increasing difference in electronegativity.

Therefore, we can say that, the greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

7 0

2 years ago