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2 years ago

A uniform rod of length l and mass m is pivoted around a line perpindicular to the rod at location l/3

Physics

1 answer:

beks73 [17]2 years ago

3 0

Answer:

I for a rod about its center is I = M L^2 / 12

A = L (1/2 - 1/3) = L / 6 distance from middle to location L/3

A^2 = L^2 / 36

I = M L^2 (1/12 + 1/36) = M L^2 (4 / 36) = M L^2 / 9

I about given location by parallel axis theorem

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Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

4 0

4 years ago

Why are fuse wires not provided in an electric circute containing an electric cell

Basile [38]

Hey there,

Electric cells we use usually carry limited voltage. Here's an example: 1.5V. Really, there is no chance of any shot circuit. that's why MCB and fuse wires are not used in their circuit .

3 0

3 years ago

An object is moving north with an initial velocity of 14 m/s accelerates 5m/s for 20 seconds. What is the final velocity of the

olga2289 [7]

Use the kinematic equation: Vf=Vi+at

Then plug;

Vi=14 m/s

a=5 m/s²

t=20 s. Therefore;

Vf=14+(5*20)

Vf=114 m/s.

6 0

4 years ago

A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air

Arte-miy333 [17]

Speed = 52/1.57576= 33 m/s

Explanation:

Distance = speed * time

Given , the distance traveled by the baseball = 52 m

Speed of sound in air = 330 m/s

Total time = 1.73333 s .

Total time for the student to hear the sound of the gong is the sum of the time take for the baseball to reach the gong and the time taken by the sound to travel back.

Distance traveled by the sound is 52 m and the speed is 330 m/s

So time taken by the sound to travel back = distance traveled / the speed

=> time = 52/330 s = 0.15757 s

Time taken by the base ball to reach the gong is the total time - the time taken by the sound

=> time taken by base ball = 1.73333 - 0.15757 = 1.57576s

Speed of the base ball to reach the gong = distance / time

Speed = 52/1.57576= 33 m/s

3 0

3 years ago

A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?

7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

$v_y=v\ sin(25)$

$v_y=22\ m/s\times\ sin(25)$

$v_y=9.29\ m/s$

Hence, the vertical component of the velocity is 9.29 m/s

3 0

4 years ago