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hammer [34]
3 years ago
10

A force of 40 N accelerates a 5 block at 6 m/s ^ 2 along a horizontal surface a. What would the block's acceleration be if the s

urface were frictionless. b. How large is the kinetic friction force? cWhat is the coefficient of kinetic friction?
Physics
1 answer:
Bas_tet [7]3 years ago
7 0

Answer:

a = 8 m/s^2, Ffriction = 10 N, μk = 0.205

Explanation:

a. Force = Mass*Acceleration,

(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)

40 N = 5 kg*acceleration,

a = 40/5 = 8 m/s^2

b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.

Fapplied - Ffriction = m*a,

40 - Ffriction = 5*6,

40 - Ffriction = 30,

Ffriction = 40 - 30 = 10 N

c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".

10 = μk*Fnormal (Fnormal = m*g = 5*9.8)

10 = μk*49,

μk=10/49 ≈ 0.205

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madam [21]

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6 0
3 years ago
Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.
kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees

ELECTRIC FORCE (F)

F = \frac{KQq}{d^{2} }

Where K = 9 x 10^{9} Nm^{2}/C^{2}

The distance between q_{1} and q_{3} can be calculated by using Pythagoras theorem.

d = \sqrt{33^{2} + 33^{2}  }

d = 46.7 cm = 0.467 m

For force F_{1}, substitute all the parameters into the formula above

F_{1} = (9 x 10^{9} x 3 x 1)/0.467^{2}

F_{1} = 2.7 x 10^{10}/0.218

F_{1} = 1.24 x 10^{11} N

For force F_{4}, substitute all the parameters into the formula above

F_{4} = (9 x 10^{9} x 3 x 4)/0.33^{2}

F_{4} = 1.08 x 10^{11}/0.1089

F_{4} = 9.92 x 10^{11} N

For force F_{2}, substitute all the parameters into the formula above

F_{2} = (9 x 10^{9} x 3 x 2)/0.33^{2}

F_{2} = 5.4 x 10^{10}/0.1089

F_{2} = 4.96 x 10^{11} N

Summation of forces on Y component will be

F_{y} = F_{4} - F_{1} Sin 45

F_{y} = 9.92 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{y} = 9.04 x 10^{11} N

Summation of forces on X component will be

F_{x} = F_{2} - F_{1} Cos 45

F_{x} = 4.96 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{x} = 4.08 x 10^{11} N

Net Force = \sqrt{F_{x} ^{2} + F_{y} ^{2}  } }

Net force = \sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2}  }

Net force = 9.9 x 10^{11} N

The direction will be

Tan ∅ = F_{y}/F_{x}

Tan ∅ = 9.04 x 10^{11} / 4.08 x 10^{11}

Tan ∅ = 2.216

∅ = Tan^{-1}(2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0
2 years ago
Read 2 more answers
A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c
Anna35 [415]

The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2

The direction of the acceleration of the ball is downwards

Learn more here: brainly.com/question/15407740

4 0
2 years ago
regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac
solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum p_{i} before the collision must be equal to the final momentum p_{f} after the collision:

p_{i}=p_{f} (1)

Being:

p_{i}=MV_{i}+mU_{i}

p_{f}=(M+m) V

Where:

M=480 g \frac{1 kg}{1000 g}=0.48 kg the mas of the peregrine falcon

V_{i}=45 m/s the initial speed of the falcon

m=240 g \frac{1 kg}{1000 g}=0.24 kg is the mass of the pigeon

U_{i}=0 m/s the initial speed of the pigeon (at rest)

V the final speed of the system falcon-pigeon

Then:

MV_{i}+mU_{i}=(M+m) V (2)

Finding V:

V=\frac{MV_{i}}{M+m} (3)

V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg} (4)

V=30 m/s (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

F=\frac{\Delta p}{\Delta t} (6)

Where:

F is the force exerted on the pigeon

\Delta t=0.015 s is the time

\Delta p is the pigeon's change in momentum

Then:

\Delta p=p_{f}-p_{i}=mV-mU_{i} (7)

\Delta p=mV (8) Since U_{i}=0

Substituting (8) in (6):

F=\frac{mV}{\Delta t} (9)

F=\frac{(0.24 kg)(30 m/s)}{0.015 s} (10)

Finally:

F=480 N

7 0
3 years ago
Drag each label to the correct location. Sort the sentences based on whether they describe the properties of a heterogeneous or
Lisa [10]

Answer:

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Explanation:

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3 years ago
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