which out of the four molecules only have a single element:Oxygen gas,liquid water,carbon disulfide,Neon gas.

What term is defined as the amount of charge stored per volt?

Shtirlitz [24]

Capacitance is a measure of charge stored per volt.

3 0

3 years ago

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b

Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $\frac{1}{2} kx^{2} +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy= $PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒ $\frac{1}{2} kx^{2} +PE1=PE2$

⇒ $PE2-PE1=\frac{1}{2} kx^{2}$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

$g$ is acceleration due to gravity=9.8 m/s

$h$ is the difference in height between two positions

⇒ $mgh=\frac{1}{2} kx^{2}$

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴ $2100*9.8*h=\frac{1}{2}*2200*0.11^{2}$

⇒ $20580*h=13.31$

⇒ $h=\frac{13.31}{20580}$

⇒h=0.0006467m= $6.5e-4$

7 0

3 years ago

Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

RUDIKE [14]

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2)v$

where

$m_1 = m_2 = m = 10 kg$ is the mass of each puck

$u_1 = 10 m/s$ is the initial velocity of the 1st puck

$u_2 = -20 m/s$ is the initial velocity of the 2nd puck

v is the final velocity of the two pucks sticking together

Re-arranging the equation and solving for v, we find:

$mu_1 + mu_2 = (m+m)v\\u_1 + u_2 = 2v\\v=\frac{u_1+u_2}{2}=\frac{10-20}{2}=-5 m/s$

Learn more about momentum:

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#LearnwithBrainly

8 0

2 years ago

When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the

Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

Voltage drop across the wire V = 0.55 volt

We know that resistance is given by $R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega$

Length of the copper wire l = 1 m

Resisitivity of the copper wire $\rho =1.72\times 10^{-8}\Omega m$

We know that resistance $R=\frac{\rho l}{A}$

$0.00177=\frac{1.72\times 10^{-8}\times 1}{A}$

$A=969.4545\times 10^{-8}m^2$

As area $A=\pi r^2$

$3.14\times r^2=969.4545\times 10^{-8}$

$r=17.57\times 10^{-4}m$

So diameter $d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m$ = 351.42 mm

3 0

3 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

2 years ago