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Sunny_sXe [5.5K]
3 years ago
8

A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introd

uced at point A, brings the block to rest at point B, 19 m to the right of point A.
What is the coefficient of kinetic friction, k , of the surface from A to B?
Physics
1 answer:
NARA [144]3 years ago
3 0

Answer:

The coefficient of kinetic friction is 0.26

Explanation:

Given:

Mass of box m = 9 kg

Initial height h = 5 m

Final height h' = 0 m

Distance travel by block d = 19 m

For finding coefficient of kinetic friction,

We use conservation laws,

Work done by frictional force is equal to change in energy,

   W_{fr} = \Delta E

Here only potential energy change so we can write,

 f_{k} d \cos 180 = mg(h') - mg(h)

Here f_{k} = \mu_{k} mg

-\mu_{k}\times mgd= -mgh

  \mu_{k} = \frac{h}{d}

  \mu _{k} = \frac{5}{19}

  \mu _{k} = 0.26

Therefore, the coefficient of kinetic friction is 0.26

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Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickl
Nikitich [7]

Answer: Relative motion

Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.

The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p

Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.

Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.

Motion of this nature is known as relative motion.

<em>Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid</em>

<em />

For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.

The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.

6 0
3 years ago
OSHA is the:
KATRIN_1 [288]
C Occupational Safety and Health Administration
8 0
1 year ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c
Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

7 0
3 years ago
A calcium bromide compound is represented by which formula?
ipn [44]

Answer:

Calcium bromide is the name for compounds with the chemical formula CaBr2(H2O)x.

4 0
3 years ago
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