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2 years ago

There is a repulsive force between two charged objects when ____________ .

1 answer:

3 0

There is a repulsive force between two charged objects when they are of like charges/ they are likely charged (like charges repel each other)

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A body with a mass of 5 Kg has the same kinetic energy as a second body. The second body has a mass of 10 kg and is moving at a

vitfil [10]

First body:

$E_{k_1}=\frac{1}{2}.m.v^2\\
\\
E_{k_1}=\frac{1}{2}.5.v_1^2=\frac{5}{2}.v_1^2$

Second body:

$E_{k_2}=\frac{1}{2}.m.v_2^2\\
\\
E_{k_2}=\frac{1}{2}.10.(20)^2=2.000 \ J$

From description of the task we have:

$E_{k_1}=E_{k_2}\\
\\
\frac{5}{2}.v_1^2=2.000 \ J\\
\\
5.v_1^2=4.000 \ J\\
\\
v_{1}^2=800 \ J\\
\\
v_{1}= \sqrt{800} \ J \approx 28,3 \ m/s$

7 0

2 years ago

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 30.0 min

schepotkina [342]

Answer:

Average speed of the trip = 52.9 km/h

Distance between initial pairs of cities (start to end) = 70.0 km

Explanation:

Since distance = speed × time

If she drives 30.0 min at 80.0 km/h

Distance covered = (30/60) × 80 = 40.0 km

Again she drives 45.0 min at 40 km/h

Distance covered = (45/60) × 40 = 30.0 km

Again she drives 12.0 min at 100 km/h

Distance covered = (12/60) × 100 = 20.0 km

Total distance covered = 40.0 + 30.0 + 20.0

= 90.0 km

Total time spent = 30.0 + 45.0 + 12.0 +15.0

= 102 min

Average speed for the trip = Total distance covered/total time spent

= 90/(102/60)

= 52.9 km/h

Distance between initial cities will be between the start of one city to the end of another

Between the first pairs = 40.0 + 30.0 = 70.0 km

Between the second pairs = 30.0 + 20.0 = 50.0 km

6 0

3 years ago

A spring, when compressed 0.20 m from the equilibrium position, stores 24 J. What is the value of the spring constant

Delvig [45]

Given that:

spring compressed ( x) = 0.2 m

Work (W) = 24 J

determine , spring constant (x)=?

We know hat ,

Work (W) = (1/2). k.x²

24 = 1/2 ( k × 0.2²)

48 = k × 0.2²

3 0

3 years ago

What is the length of a one-dimensional box in which an electron in the n=1 state has the same energy as a photon with a wavelen

Pie

Answer:

Length will be 0.491 nm

Explanation:

We have given wavelength of the photon $\lambda =800nm=800\times 10^{-9}m$

Plank's constant $h=6.6\times 10^{-34}J-s$

We know that energy of the photon is given by

$E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{800\times \times 10^{-9}}=2.475\times 10^{-19}J$

We know that energy of photon is also given by

$E=\frac{n^2h^2}{8mL^2}=\frac{h^2}{8mL^2}$

$2.475\times 10^{-19}=\frac{(6.6\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times L^2}$

$L=0.491\times 10^{-9}m$

3 0

3 years ago

Four identical charges, Q, occupy the corners of a square with sides of length a. A fifth charge, q, can be placed at any desire

Nat2105 [25]

Answer:

q = - ( 2*sqrt(2) + 1 )*Q / 4

Explanation:

Given:

- Side of each square L = a

- Charge q is placed among 4 other identical charges Q.

Find:

- Find the location and magnitude of the fifth charge q such that net Electric Force at its position is zero.

Solution:

- Compute distance r from charge Q to q i.e center of all four charges Q:

r = 0.5*sqrt ( a^2 + a^2 )

r = a / 2sqrt(2)

- Compute the individual Electrostatic forces @ point A:

F_b = F_d = k*Q^2 / a^2

F_c = k*Q^2 / (a*sqrt(2))^2 = k*Q^2 / 2*a^2

F = k*Q*q / (a / 2sqrt(2))^2 = 2*k*Q^2 / a^2

- Use Electrostatic Equilibrium conditions:

F_b + F_c*cos(45) = F*cos(45)

F_d + F_c*sin(45) = F*sin(45)

- Plug in the values and equate:

{ (Q/a)^2 + (Q^2 / 2*a^2*sqrt(2)) = sqrt(2)*Q*q / a^2

- Canceling all k's and a^2:

Q * ( 1 + (1 /2*sqrt(2)) ) = sqrt(2)*q

q = - ( 2*sqrt(2) + 1 )*Q / 4

- Note: In attachment Q's and q's are interchange but the solution here provided is according to the question at hand.

3 0

3 years ago