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FromTheMoon [43]
3 years ago
12

Calculate the kinetic energy of a 1 kg object moving at 4.5 ms. Round to the first decimal.

Physics
1 answer:
harina [27]3 years ago
8 0

Answer:

91.21

Explanation:

Kinetic energy =1/2 mv2 substitute the values I. E 1/2 *1 *4.5^2= 91.21

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A rifle fires a 2.90 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the
iren2701 [21]

Answer: 586.60N/m

Explanation:

In this scenario, the elastic potential energy of the spring is converted into potential energy.

0.5*K*x^2 = mgh

Thus K = 2mgh/x^2

=(2*2.90*10^-2*9.8*7.23)/(8.37*10^-2)^2

=586.599

Therefore K = 586.60N/m

7 0
4 years ago
Read 2 more answers
The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf
kolezko [41]

Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy  

V0x = V0 cos(θ) 

V0y = V0 sin(θ) 

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2 

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant  

Vx = Vocos(θ) 

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t 

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t 

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2 

3 0
4 years ago
Which two statements are true about a system?
shusha [124]
A and c

Because it defines the boundaries of the system
4 0
3 years ago
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Question 15: At the edge of a lake you throw a stone with a velocity of 29 m/s at an angle of 45°. The stone is in the air for 4
Marianna [84]
Get your numbers gathered up and solve the problem in the ordered step
8 0
3 years ago
A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.
Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence  ( 65.0° )

{\theta_r} is the angle of refraction  ( 53.0° )

{n_r} is the refractive index of the refraction medium  (unknown liquid, n=?)

{n_i} is the refractive index of the incidence medium (oil, n=1.38)

Hence,  

1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}

Solving for {n_r},

Refractive index of unknown liquid = 1.56

4 0
4 years ago
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