PHYSICS COLLEGE

A bug flying horizontally at 1.7 m/s collides and sticks to the end of a uniform stick hanging vertically from its other end. After the impact, the stick swings out to a maximum angle of 7.0Â° from the vertical before rotating back. If the mass of the stick is 16 times that of the bug, calculate the length of the stick (in m).

Answers

Answer 1

Answer:

Answer:12.11 m

Explanation:

Given

Bug speed =1.7 m/s

Let mass of bug is m

mass of rod 16m

maximum angle turned by rod is 7^{\circ}[/tex]

From Energy conservation

kinetic energy of bug =Gain in potential energy of rod

$(1)/(2)mv^2=16mgL(1-cos\theta )$

$L=(1.7^2)/(2* 16(1-cos7))$

L=12.11 m

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A 55-liter tank is full and contains 40kg of fuel. Find using Sl units: • Density p. • Specific Weight y • Specific Gravity Answer tolerance = 1%. Be sure to include units. The sign of the answers will not be graded, use a positive value for your answer. Your answers: p= (Enter a positive value) y = (Enter a positive value) SG = (Enter a positive value)

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answers

To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_(1,2)$ = Mass of each object

$v_(1,2)$ = Initial Velocity of each object

$v_f$ = Final velocity

Replacing we have that,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

$1850*13.8+3100*0 = (1850+3100)v_f$

$v_f = 5.1575m/s$

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

$v_f^2-v_i^2 = 2ax$

Since there is no initial speed, then

$v_f^2 = 2ax$

$5.1575^2 = 2a (1.91)$

$a = 6.9633m/s^2$

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

$F_f = F_a \n\mu N = m*a \n\mu = (ma)/(N)\n\mu = (ma)/(mg)\n\mu = (a)/(g)\n\mu = (6.9633)/(9.8)\n\mu = 0.7105$

Therefore the Kinetic friction coefficient is 0.7105

Some snakes have special sense organs that allow them to see the heat emitted from warm blooded animals what kind of an electromagnetic waves does the sense organs detect?A. Visible light waves
B. Ultraviolet light waves
C. Infrared Waves
D. Microwaves

Answers

The heat emitted from anything is carried in the form of infrared waves. (C)

Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?

Answers

Answer:

128 s

Explanation:

The distance, speed and time are related as;

$Distance=Speed* Time$

Given that the speed = 5 m/s

Distance = 640 m

Time = ?

So,

$Distance=Speed* Time$

$640\ m=5\ m/s* Time$

$Time=\frac {640\ m}{5\ m/s}=128\ s$

Thus, Garza takes 128 s to travel 640 m at 5 m/s speed.

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five times smaller

Answers

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.

Now, let's consider moment of inertia = I and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia = (five times larger)

angular speed = ω/5 (five times smaller)

The kinetic energy of a spinning body is given as:

$K.E.=(1)/(2) I. w^2$

On substituting the values, we will get:

$K.E.= (1)/(2) (5I) ((w)/(5) )^2 \n\nK.E. =(1)/(10) I. w^2$

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

Learn more:

brainly.com/question/12337396

A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.

Answers

Answer:

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance = 100 cm

v = Image distance

f = Focal length = -23.9 cm (concave lens)

$h_u$ = Object height = 2.1 cm

Lens Equation

$(1)/(f)=(1)/(u)+(1)/(v)\n\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\n\Rightarrow (1)/(v)=(1)/(-23.9)-(1)/(100)\n\Rightarrow (1)/(v)=(-1239)/(23900) \n\Rightarrow v=(-23900)/(1239)=-19.29\ cm$

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

Magnification

$m=-(v)/(u)\n\Rightarrow m=-(-19.29)/(100)\n\Rightarrow m=0.1929$

$m=(h_v)/(h_u)\n\Rightarrow 0.1929=(h_v)/(2.1)\n\Rightarrow h_v=0.1929* 2.1=0.40509\ cm$

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?