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maw [93]
3 years ago
9

Scientific question must be

Physics
1 answer:
Rama09 [41]3 years ago
8 0
Are there any choices with the question?
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What is another unit for momentum besides kg-m/s?<br> a. N<br> b. N-s<br> c. N-s2<br> d. N/s
lara [203]
'Newton-second' is dimensionally equivalent to 'kilogram-meter/second'.
8 0
3 years ago
A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius
disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

  • Disk radius R = 15 cm = 0.15 m
  • Cylinder radius r = 1.5 cm = 0.0015 m
  • Disk mass M = 20 kg
  • Time t = 1.2 s
  • Force F = 12 N

to find

  • Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

        KE = \frac{F^2 \ R^4}{M \ R^2 } \ t^2

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

4 0
3 years ago
PLEASE HELP!!! WILL GET BRAINIEST FOR 5 SENTENCES
emmasim [6.3K]
Hello.

BEVs and hydrogen fuel cell vehicles are a more promising transportation technology for the future because they reduce greenhouse gas emissions as well as CO2 making it more 'green.' The second question is your opinion, in mine no, they should not be required but they should at least be considering.
Have a nice day
5 0
3 years ago
Collapse question part Part 4 (d) What is the unit vector in the direction of the spacecraft's velocity? (Express your answer in
Maksim231197 [3]

Answer:

unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

Explanation:

Given:

                            v =  (-23.2, -104.4, 46.4) m/s

Above expression describes spacecraft's velocity vector v.

Find:

Find unit vector in the direction of spacecraft velocity v.

Solution:

Step 1: Compute magnitude of velocity vector.

                            mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)

                            mag (v) = 116.58 m/s

Step 2: Compute unit vector unit (v)

                            unit (v) = vec (v) / mag (v)

                            unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58

                            unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]

7 0
3 years ago
The separation in time between the arrival of primary and secondary waves is
Gre4nikov [31]
The separation in time between the arrival of primary and secondary wave is called LAG TIME.
The time difference between the arrival of primary wave and secondary wave in a seismogram is called lag time. The primary wave always travels faster than the secondary wave, thus the difference between the two can be obtained by estimating the difference between the arrival time of the two waves/.
3 0
3 years ago
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