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2 years ago

What is the kinetic energy of a 100 kg object that is moving with a speed of 12.5m/s

Physics

1 answer:

Doss [256]2 years ago

7 0

The kinetic energy of any moving object is

(1/2) (mass) (speed²) .

For the object you described, that's

(1/2) (100 kg) (12.5 m/s)²

= (50 kg) (156.25 m²/s²)

= 7,812.5 joules
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A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

2 years ago

Isabella deja caer accidentalmente un bolígrafo desde su balcón mientras celebra que resolvió satisfactoriamente un problema de

jok3333 [9.3K]

Answer:

i cant read spanish lol

Explanation:

8 0

1 year ago

Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache

FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis

N - W cos θ = 0

The equation for the force of friction is

fr = μ N

fr = μ mg cos θ

We replace at work

W = - μ m g cos θ d

Mechanical energy in the lower part of the embankment

Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

$Em_{f}$ = U = m g y

W = ΔEm = $Em_{f}$ - Em₀

- μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

sin θ = y / d

y = d sin θ

- μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

d (g syn θ + μ g cos θ) = ½ v²

d = v²/2 g (sintea + myy cos tee)

d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

d = 1555.85 /7.8145

d = 199.1 m

Let's use trigonometry to find the height

sin 16 = y / d

y = d sin 16

y = 199.1 sin 16

y = 54.9 m

8 0

2 years ago

The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

Lisa [10]

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J

3 0

2 years ago

Read 2 more answers

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

ankoles [38]

Answer:

The magnitude of the centripetal acceleration during the turn is $a=12.04\ m/s^2.$

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , $T=t\times 4=60\ s.$

Now , Velocity ,

$v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.$

Also , we know :

Centripetal acceleration ,

$a=\dfrac{v^2}{r}$

Putting all values we get :

$a=12.04\ m/s^2.$

Hence , this is the required solution .

5 0

1 year ago