Plzzzzzzzzzzzzz helppppp 20 points

A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t

8090 [49]

Answer:

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass c.m= 1.6m

hence 1kg is 1.6m from the c.m

and x is 0.4m from the c.m

Taking moment about the c.m

clockwise moment equals anticlockwise moments

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

3 0

3 years ago

Water is pumped through a pipe of diameter 15.0 cm, from the Colorado River up to Grand Canyon Village, located on the rim of th

Aleksandr [31]

Answer:

p= 1.50289×10⁷ N/m²

Explanation:

Given

HA = (564 m)................(River Elevation)

HB = (2096 m).............(Village Elevation)

Area = A =(π/4){Diameter}² = (π/4){0.15 m}² = 0.017671 m²

ρ = (1 gram/cm³) = (1000 kg/m³)........(Water Density)

p(pressure)=?

Solution

p=PA - PB

p= ρ*g*HB - ρ*g*HA

p= (ρ*g)*(HB - HA)

p= (1000×9.81 )×{2096 - 564}

p= 1.50289×10⁷ N/m²

8 0

3 years ago

Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?

morpeh [17]

Answer: In a longitudinal wave, the crest and trough of a transverse wave correspond respectively to the compression, and the rarefaction. A compression is when the particles in the medium through which the wave is traveling are closer together than in its natural state, that is, when their density is greatest.

4 0

2 years ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

6 0

3 years ago

A CO molecule has an intrinsic dipole moment whose magnitude is 4 X 10^-31 C*m. If the separation between the atoms is 0.11 nm,

aleksandrvk [35]

4x10^-31/(0.11x10^-9)= 3,63 x 10^-21

5 0

3 years ago