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sashaice [31]
3 years ago
15

You pull with a force of 255 N on a piece of luggage of mass 68 kg, but it does not move. If the coefficient of friction between

the luggage and the floor is 0.7, what is the force of static friction acting on the luggage?
Physics
2 answers:
VLD [36.1K]3 years ago
8 0

Answer:

the other guy was right, the answer is 255

Explanation:

STALIN [3.7K]3 years ago
3 0

The force of static friction acting on the luggage =255 N

Explanation:

Applied force= 255 N

mass of luggage=68 kg

coefficient of friction= μ=0.7

Normal force=N = mg

N= 68(9.8)=666.4 N

now the force of friction is given by

Ff= μN

Ff=0.7 (666.4)

Ff=466.5 N which is greater than the applied force.

When the applied force is less than the force of friction, in that case

force of friction= applied force

Thus the force of station friction acting on the luggage = 255 N

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A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engin
Sidana [21]

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

8 0
3 years ago
n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us
barxatty [35]

Answer:

The answer is "(1.265 \pm 0.010) \ s \ and \ 0.709 \%"

Explanation:

In point i:

T_{theo}= 2\pi \sqrt{\frac{l}{g}}

        =2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s

If  error in the theoretical time period :

\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2}  \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}

           = 0.010 \ s

 T_{theo} = (1.265 \pm 0.010) \ s

In point ii:

\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100

<h3>                     = \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%</h3>
5 0
3 years ago
A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl
USPshnik [31]

Given Information:  

Wavelength =  λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0
3 years ago
A woman does 236 J of work
GREYUIT [131]

Answer:

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3 0
3 years ago
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