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2 years ago

7

For the same setup as in the previous question, what is the rotational kinetic energy of the spinning cube after the string unwi

nds L=0.30m? Use Energy methods. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Be careful with this problem. Both the spinning cube picks up kinetic energy as it spins up and the hanging mass gains kinetic energy as the string unwinds. Rotational kinematics will be necessary e.g.

1 answer:

bezimeni [28]2 years ago

8 0

Answer:

Kinetic energy = 0.145 J

Explanation:

See the attached file for explanation.

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The velocity of a an object in linear motion changes from +25 meters per second to +15 meters per second in 2.0 seconds.

kodGreya [7K]

Answer:

$-5.0 m/s^2$

Explanation:

Missing question:

What is the object's acceleration?

Solution:

The acceleration of an object is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the change in velocity

For the object in this problem,

u = +25 m/s

v = +15 m/s

t = 2.0 s

Substituting,

$a=\frac{15-25}{2}=-5.0 m/s^2$

And the acceleration is negative because its direction is opposite to that of the velocity.

5 0

2 years ago

The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1

Andrei [34K]

Answer:

The expression of gravitational field due to mass $m_!$ at a distance $l_a$

Explanation:

We have given mass is $m_1$

Distance of the point where we have to find the gravitational field is $l_a$

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by $g=\frac{Gm_1}{l_a^2}$

This will be the expression of gravitational field due to mass $m_!$ at a distance $l_a$

4 0

2 years ago

A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured �� – �� with respect to the positi

Komok [63]

Answer:

223 degree

Explanation:

We are given that

Magnitude of resultant vector= 8 units

Resultant vector makes an angle with positive -x in counter clockwise direction

$\theta=43^{\circ}$

We have to find the magnitude and angle of the equilibrium vector.

We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.

Therefore, magnitude of equilibrium vector=8 units

x-component of a vector= $v_x=vcos\theta$

Where v=Magnitude of vector

Using the formula

x-component of resultant vector= $v_x=8cos43=5.85$

y-component of resultant vector= $v_y=vsin\theta=8sin43=5.46$

x-component of equilibrium vector= $v_x=-5.85$

y-component of equilibrium vector= $-v_y=-5.46$

Because equilibrium vector lies in III quadrant

$\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}$

The angle $\theta'$ lies in III quadrant

In III quadrant ,angle = $\theta'+180^{\circ}$

Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree

4 0

2 years ago

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

$Y =\frac{mgL}{A\Delta L}$

$\Delta L =\frac{mgL}{A\times Y}$

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0

2 years ago

1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and

saul85 [17]

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

Centripetal acceleration = $\frac{v^{2} }{r}$

v is the velocity of the body

r is the radius of the circle

putting in the parameters:

Centripetal acceleration = $\frac{4^{2} }{0.75}$

Centripetal acceleration = 10.67m/s²

Centripetal force = m $\frac{v^{2} }{r}$

m is the mass

Centripetal force = mass x centripetal acceleration

= 3 x 10.67

= 32N

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

4 0

2 years ago