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prisoha [69]
4 years ago
15

Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the distance separating objects 1 and 2 is chang

ed to one-half the original value (I.E halved), then the new electrostatic force will be _____ units

Physics
2 answers:
qaws [65]4 years ago
5 0

Answer:

288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.

Explanation:

If all other parameters are constant,

Electrostatic Force of attraction ∝ (1/r²)

F = (k/r²) = 72.0

If r₁ = r/2, what happens to F₁

F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units

Lemur [1.5K]4 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t
Neko [114]

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

v_f^2 - v_i^2 = 2 a d

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

v_i = 0

also the acceleration of the stone is due to gravity which is given as

g = 32 ft/s^2

now we have

v_f = 180 ft/s

so from above equation

180^2 - 0 = 2(32)d

d = 506.25 ft

6 0
3 years ago
A large refrigerator (mass 80kg) sits at rest inside a house. The homeowner wants to move the refrigerator across the room, so s
Liono4ka [1.6K]

Answer:

400 N

Explanation:

By the law of friction,

F=\mu R

F is the maximum frictional force, \mu is the coefficient of friction and R is the reaction on the refrigerator. On a horizontal surface, the reaction is equal to the weight of the refrigerator.

R=mg

F=\mu mg

While not moving, the fricition on the refrigerator is static friction. So, \mu=0.65

F=0.65 \times80\times9.8=509.6 \text{ N}

This is the maximum frictional force and is more than the applied horizontal force of 400 N. Frictional force cannot be more than the applied force, else it would actually pull the refrigerator backwards (a strange thing, if it were to happen). It is equal to the extent of the applied force because the applied force is not enough to overcome the maximum.

Hence the frictional force is 400 N.

PS: Note that we do not use the coefficient of kinetic friction because applied force could not overcome the static friction.

6 0
4 years ago
A 1000 kg car pushes a 2000 kg car that has a dead battery. When the driver steps on the accelerator, the drive wheels of the ca
butalik [34]
So what is the question
4 0
3 years ago
I have a question, it concerns hydrostatic buoyancy and Archimedes' law.
saw5 [17]

Answer: the lvl wud remain the same

Explanation: as per Archimedes Principle, the weight of the water displaced by the object is equal to the weight of the object. When the ship initially went into the pool, it wud hv displaced some water. When the anchor is dropped, the level does not change coz the anchor was already in the ship and no extra weight has been added, so the weight of the anchor has already been accounted for in the first place when the ship was first placed in the pool

4 0
3 years ago
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis
ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0
4 years ago
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