What is the speed of a transverse wave in a rope of length 2 meters and mass 0.06

What would make oppositely charged objects attract each other more? increasing the positive charge of the positively charged obj

kykrilka [37]

Answer:

A.

Explanation:

I did the test :D

3 0

3 years ago

Read 2 more answers

PLEASE HELP

hichkok12 [17]

The correct answer is 195.6 N

Explanation:

Different from the mass (total of matter) the weight is affected by gravity. Due to this, the weight changes according to the location of a body in the universe as gravity is not the same in all planets or celestial bodies. Moreover, this factor is measured in Newtons and it can be calculated using this simple formula W (Weight) = m (mass) x g (force of gravity). Now, leps calculate the weigh of someone whose mass is 120 kg and it is located on the moon:

F = 120 kg x 1.63 m/s2

F= 195.6 N

7 0

3 years ago

if you have 34 pounds of meat at 200F and you immerse it in 15 liters of water at 25C what is the final temp of the water? ( spe

hram777 [196]

Answer:

The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT ⨉ m)

5 0

3 years ago

Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at 3.2 MW. What is the intensity of

Dahasolnce [82]

Answer:

$I=1.5\times10^{-28}W/m^2$

Explanation:

The intensity is related to the power and surface area by $I=\frac{P}{A}=\frac{P}{4\pi r^2}$ . We need to calculate the surface area of a sphere of radius r=4.3ly.

Since 4.3ly is the distance light travels in 4.3 years at 299792458m/s, we can obtain it in meters by doing:

$r=vt=(299792458m/s)(4.3\times365\times24\times60\times60s)=4.1\times10^{16}m$

So we have:

$I=\frac{P}{4\pi r^2}=\frac{3.2\times10^6W}{4\pi (4.1\times10^{16}m)^2}=1.5\times10^{-28}W/m^2$

4 0

3 years ago

How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners

Natasha_Volkova [10]

Answer:

$Potential\ Energy=Work \ Done=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V= $\frac{kq}{r}$

So,

$Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}$

Where:

k is Coulomb Constant= $8.99*10^9 Nm^2/C^2$

q is the charge on electron= $-1.6*10^-19 C$

r is the distance= $3.0*10^{-10}m$

For 3 Electrons Potential Energy or work Done is:

$Potential\ Energy=3*\frac{kq^2}{r}$

$Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

7 0

3 years ago