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nika2105 [10]
3 years ago
14

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

Physics
1 answer:
aev [14]3 years ago
5 0

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

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A person lifts a 25 kg box of old sports equipment at an angle of 75 degrees to the vertical to a shelf 2.6 meters above. How mu
kogti [31]

Answer:

164.87 J

Explanation:

From the question,

Work done (W) = mghcosθ........................ Equation 1

Where m = mass of the box, h = height, g = acceleration due to gravity, θ = angle to the vertical

Given: m = 25 kg, h = 2.6 meters, θ = 75°.

Constant: g = 9.8 m/s²

Substitute these value into equation 1

W = 25×9.8×2.6×cos75°

W = 164.87 J.

6 0
3 years ago
Car a with mass 1,783 kg collides with stationary 1600 kg car b. they become locked together after the collision and move with s
ValentinkaMS [17]

The initial speed of car A is 15.18 m/s.

Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.

p=p'

pa+pb = pa'+pb'

(ma×va) + (mb×vb) = (ma×va') + (mb×vb')

  • ma = mass of object A (kg) = 1,783 kg
  • mb = mass of object B (kg) = 1,600 kg
  • va = speed of object A before collides (m/s)
  • va' = speed of object A after collides (m/s) = 8 m/s
  • vb = speed of object B before collides (m/s) = 0 m/s
  • vb' = speed of object B after collides (m/s) = 8 m/s
  • p = momentum before collision (Ns)
  • p' = momentum after collision (Ns)

(ma×va) + (mb×vb) = (ma×va') + (mb×vb')

(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)

(1,783×va) + 0 = 14,264+12,800

(1,783×va) = 27,064

va \:=\: \frac{27,064}{1.783}

va = 15.18 m/s

Learn more about The law of momentum conservation here: brainly.com/question/7538238

#SPJ4

3 0
1 year ago
A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn
vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted  by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

f_{m} = \mu_{s}N

where

N = Normal reaction = mg

Thus

f_{m} = \mu_{s}mg

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

f\leq f_{m}

f \leq \mu_{s}mg

109 \leq \mu_{s}\times 31.2\times 9.8

\mu_{s}\geq 0.356

7 0
3 years ago
Kinetic energy can be passed from one object to another when objects collide,
antiseptic1488 [7]
Kinetic energy can be passed from one object to another when objects collide,
Answer: True

Hope This Helps! :3
5 0
3 years ago
Read 2 more answers
A race is held between a sports car and a motorcycle. The sports car can accelerate at 5.0 m/s^2 and the motorcycle can accelera
user100 [1]

Answer:

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

To find the speed of car after 5 s, we use 1st equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car = ?

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

<u>Vf₁ = 25 m/s</u>

<u></u>

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

t₂ = time = 6 s

Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

<u>s₂ = 90 m</u>

<u></u>

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

t₁ = √60 s²

t₁ = 7.74 s

For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₁ = 200 m

Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.

5 0
3 years ago
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