Work is the product of ___<br> _ and an object's displacement.

A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height

stira [4]

Answer:

$9.5 kg m^2/s$

Explanation:

The angular momentum of an object is given by:

$L=mvr$

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

$L=(0.35)(9.0)(3.0)=9.5 kg m^2/s$

4 0

3 years ago

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at

Vesna [10]

Answer:

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0 So B=0

-16 B-32 A+16B=6 So A=-3/16

y=-3/16 cos 4t

Now to find the CF of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

$m^2+8m+16=0$

So $CF =(C_1+tC_2)e^{-2t}$

So the general equation

$Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t$

Given that t=0 Y=0 So

$C_1=\dfrac{3}{16}$

t=0 Y'=0 So

$C_2 =\dfrac{3}{8}$

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

The above equation is the general equation for motion.

3 0

3 years ago

How are the mass and weight of an object related? Include a description with words and an equation.

nikdorinn [45]

The weight is the force experienced, whereas the mass represents the actual quantity of matter inside a body..
weigh on the surface of the earth is equal to mg
mass is m
and at the centre weight is 0 due to 0 acceleration that's 0 g
but mass is always constant and remains m, no matter where you are

6 0

3 years ago

Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?

OverLord2011 [107]

Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

5.0 min × (60 s / min) = 300 s

Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

3 0

2 years ago

A 2 feet piece of wire is cut into two pieces and once a piece is bent into a square and the other is bent into an equilateral t

SpyIntel [72]

Answer:

Explanation:

Let x ft be used to make square and 2-x ft be used to make equilateral triangle.

each side of square = x/4

area of square = ( x /4 )²

Each side of triangle

= (2-x) /3

Area of triangle = 1/2 (2-x)²/9 sin 60

= √3 / 36 x (2-x)²

Total area

A = ( x /4 )² +√3 / 36 (2-x)²

For maximum area

dA/dx = 0

1/16( 2x ) -√3 / 36 x2(2-x) = 0

x / 8 - √3(2-x)/ 18 = 0

x / 8 - √3/9 + √3/18 x = 0

x ( 1/8 + √3/18 ) = √3/9

x(.125 +.096 ) = .192

x = .868 ft

3 0

2 years ago

Work is the product of ___ _ and an object's displacement.