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Paha777 [63]
3 years ago
12

Feeling slightly uncomfortable during exercise

Physics
1 answer:
icang [17]3 years ago
8 0
A it is completely normal and is to be expected. 


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A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin
Goshia [24]

Answer:

(A) 0.63 J  

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + mh^{2}  

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5Iω^{2}

  = 0.5 x 0.07875 x 4^{2} = 0.63 J   0.15 m

(B) from the conservation of energy

   initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

   Ki + Ui = Kf + Uf

   at the maximum height velocity = 0 therefore final kinetic energy = 0

   Ki + Ui = Uf

   Ki = Uf - Ui

 Ki =  mg(H-h)

where (H-h) = rise in the center of mass

     0.63 = 0.42 x 9.8 x (H-h)

   (H-h) = 0.15 m

6 0
3 years ago
Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in the collisio
Dmitriy789 [7]

During the collision between two balls on the pool table there is no external force along the line of collision between them

Since there is no external force on it so here we can say

F = 0 = \frac{\Delta P}{\Delta t}

here we have

\Delta P = 0

so we can say

P_i = P_f

since there is no external force so we can say during the collision the momentum of two balls will remain conserved

7 0
3 years ago
Read 2 more answers
HELP DUE IN 5 How does a battery supply energy to a circuit?
Gnoma [55]

Answer: It's C

Explanation:

3 0
2 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
If you were to travel from the equator to the higher latitudes (near either pole) on Saturn, the differential rotation would cau
Dmitrij [34]

Answer:

Saturn's differential rotation will cause the length of a day measures to be longer by 0.4 hours

Explanation:

Differential rotation occurs due to the difference in angular velocities of an object as we move along the latitude of the or as we move into different depth of the object, indicating the observed object is in a fluid form

Saturn made almost completely of gas and has differential motion given as follows

Rotation at the equator = 10 hours 14 minutes

Rotation at high altitude = 10 hours 38 minutes

Therefore;

The differential rotation = 10 hours 38 minutes - 10 hours 14 minutes

The differential rotation = 24 minutes = 24 minutes × 1 hour/(60 minutes) = 0.4 hours

The differential rotation = 0.4 hours

Therefore, the measured day at the higher altitude will be 0.4 longer than at the equator.

8 0
3 years ago
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