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lara31 [8.8K]
1 year ago
13

Suppose that a simple pendulum consists of a small 60.0 g bob at the end of a cord of negligible mass. If the angle 0 between th

e cord and the vertical is given by g - (0.0800 rad) cos[(4.43 rad/s) t 0], what are (u) the pendulum's length and (b) its maximum kinetic energy
Physics
1 answer:
erik [133]1 year ago
5 0

Based on the mass of the bob and the angle between the cord and the vertical, the pendulum length is 0.50m.

The maximum kinetic energy can be found to be 9.42 x 10⁻⁴J.

<h3>What is the pendulum length?</h3>

This can be found as:

= g-force / w²

Solving gives:

= 9.8 / 4.43²

= 0.4998 m

= 0.50 m

<h3>What is the maximum kinetic energy?</h3>

This can be found as:

= 0.5 × m × w² × A²

Maximum kinetic energy is:

= 0.5 × 60 × 10⁻³ × (4.43 × 0.4998 x 0.08 rad)²

= 9.42 x 10⁻⁴J

Find out more on maximum kinetic energy at brainly.com/question/24690095.

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Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment
Drupady [299]

Answer:

u_K=0.862

Explanation:

The force of friction between the quails feet and the ground is:

F=m*a

F_K=m*a

F_K=u_k*m*g

u_K*m*g=m*a_c

u_K*g=a

u_K=\frac{a_c}{g}

a_c=\frac{v^2}{r}

So the coefficient of static is solve

u_K=\frac{\frac{v^2}{r}}{g}

u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}

u_K=0.862

4 0
3 years ago
Give THREE examples of objects that have potential energy
Tasya [4]

A ball kept on 3rd floor of a building.

A pendulum bob kept at 3m height

A stone thrown vertically upward.

A pressed spring.

A squashed spunge ball.

4 0
2 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg
Yanka [14]
Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
6 0
3 years ago
Read 2 more answers
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

  • f is the focal length of the mirror
  • v is the object distance
  • u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

#SPJ1

7 0
1 year ago
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