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goblinko [34]
3 years ago
10

(i) a force of 35.0 n is required to start a 6.0-kg box moving across a horizontal concrete floor, (a) what is the coefficient o

f static friction between the box and the floor? (b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2. what is the coefficient of kinetic friction?
Physics
1 answer:
Serjik [45]3 years ago
4 0

a. The force applied would be equal to the frictional force.

F = us Fn

where, F = applied force = 35 N, us = coeff of static friction, Fn = normal force = weight

 

35 N = us * (6 kg * 9.81 m/s^2)

us = 0.595

 

b. The force applied would now be the sum of the frictional force and force due to acceleration

F = uk Fn + m a

where, uk = coeff of kinetic friction

 

35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)

uk = 0.533

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The mass of the aeroplane is 300,000 kg.

<h3>What is Newton's second law of motion?</h3>

It states that the force F is directly proportional to the acceleration a of the body and its mass.

The law is represented as

F =ma

where acceleration a = velocity change v / time interval t

Given is the aeroplane lands at a speed of 80 m/s. After landing, the aeroplane takes 28 s to decelerate to a speed of 10 m/s. The mean resultant force on the aeroplane as it decelerates is 750 000 N.

The force expression will be

F = mv/t

Substitute the values and we have

750000 = m x  (80 -10)/ 28

750,000 = m x 2.5

m = 300,000 kg

Thus, the mass of the aeroplane is 300,000 kg.

Learn more about Newton's second law of motion.

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2 years ago
HELP!!! 30 POINTS+BRAINLIEST!!!!!QUICK!!
tia_tia [17]

Answer:

A:7.2ms^{-1}

B:14.25ms^{-1}

C:1.45sec

D:10.3m

E:2.9sec

F:20.88m

Explanation:

Let v be the velocity and \alpha be the angle between the velocity and ground.

Question A:

Horizontal component of velocity is given by vcos(\alpha ).

So,horizontal component is 16\times cos(63)=16\times 0.45=7.2ms^{-1}

Question B:

Vertical component of velocity is given by vsin(\alpha ).

So,vertical component is 16\times sin(63)=16\times 0.89=14.25ms^{-1}

Question C:

Time required is given by \frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds

Question D:

Maximum height is given by \frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m

Question E:

Time of flight is twice the time required to reach maximum height=2\times 1.45=2.9 seconds.

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The distance between the player and ball after landing is called range and is given by \text{horizontal component of velocity}\times \text{time of flight}=7.2\times 2.9=20.88m

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Is a personal trainer considered a doctor?
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3 years ago
An archer pulls her bowstring back 0.396 m by exerting a force that increases uniformly from zero to 237 N. (a) What is the equi
tatyana61 [14]

Answer:

(a) The equivalent spring constant is 598.485 N/m

(b) The work done is 46.926 J

Explanation:

From Hooke's law of elasticity

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K = F/e = 237/0.396 = 598.485 N/m

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