All categories

2 years ago

Please helppppppp!!!!!!

Physics

1 answer:

Mamont248 [21]2 years ago

6 0

Answer:

The resulting force on the first object is 800 N.

Explanation:

Given;

force exerted on one of the objects, F₁ = 400 N

let the first charge = q₁

let the second charge = q₂

The force of repulsion between the objects is calculated using Coulomb's law;

$F =\frac{kq_1q_2}{r^2} \\\\\frac{k}{r^2} = \frac{F}{q_1q_2} = \frac{F_1}{q_1\times2q_2} \\\\F_1 = \frac{F(q_1\times2q_2)}{q_1q_2} \\\\F_1 = 2F\\\\F_1 = 2(400 \ N)\\\\F_1 = 800 \ N$

Therefore, the resulting force on the first object is 800 N.

You might be interested in

The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr

ch4aika [34]

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring

$\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}$

$2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2$

$v_2=3.67 m/s$

7 0

3 years ago

How much energy is required to move an electron through a potential difference of

Tresset [83]

I think it’s going to be the 2nd one

3 0

2 years ago

Write the SI unit of time and temperature

dsp73

Answer:

The SI unit of time is second (s) and temperature is Kelvin (K)

Explanation:

hope it is helpful to you

7 0

3 years ago

What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i

babymother [125]

Answer:

$\Delta S=1.69J/K$

Explanation:

We know,

$\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$ ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

$0.28=1-\frac{Q_2}{3.78\times 10^{3}}$

$\frac{Q_2}{3.78\times 10^3}=0.72$

$Q_2=3.78\times 10^3\times0.72$

⇒ $Q_2 =2.721\times 10^3 J$

Now,

The entropy change ( $\Delta S$ ) is given as:

$\Delta S=\frac{\Delta Q}{T_1}$

$\Delta S=\frac{Q_1-Q_2}{T_1}$

substituting the values in the above equation we get

$\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}$

$\Delta S=1.69J/K$

7 0

2 years ago

The acronym laser stands for light amplification by ____ emission of radiation

AleksAgata [21]

<span>light amplification by stimulated emission of radiation </span>

6 0

3 years ago