The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (1,127 kg) x (6 m/s² forward)
= (1,127 x 6) newtons forward
= 6,762 newtons forward
______________________________
Momentum = (mass) x (speed)
= (69 kg) x (6 m/s)
= 414 kg-m/s
Answer:
<em>J=36221 Kg.m/s</em>
Explanation:
<u>Impulse-Momentum Theorem</u>
These two magnitudes are related in the following way. Suppose an object is moving at a certain speed
and changes it to
. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

The initial and final momentums are, respectively

The change of momentum is

It is numerically equal to the Impulse J


We are given

The impulse the car experiences during that time is

J=-36221 Kg.m/s
The magnitude of J is
J=36221 Kg.m/s
In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.
The equation that defines the linear moment is given by

where,
m=Total mass
Mass of Object
Velocity before throwing
Final Velocity
Velocity of Object
Our values are:

Solving to find the final speed, after throwing the object we have

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.
That way during each section the equations should be modified depending on the previous one, let's start:
A) 



B) 



C) 



Therefore the final velocity of astronaut is 3.63m/s
The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E







Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
Learn more about angular velocity here:
brainly.com/question/1452612