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antiseptic1488 [7]

2 years ago

5

A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitca

se slides 2.0 m before stopping. Use work and energy to find the suitcase’s coefficient of kinetic friction on the floor.

1 answer:

Hatshy [7]2 years ago

7 0

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

$u=\frac {v^{2}}{2gd}$

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

$u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367$

Therefore, the coefficient of kinetic friction is approximately 0.0367

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Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

Learn more about friction:

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#LearnwithBrainly

7 0

2 years ago

A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o

faltersainse [42]

m = mass = 5 kg

$v_{i}$ = initial velocity = 100 m/s

$v_{f}$ = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m( $v_{f}$ - $v_{i}$ )

30 = 5 ( $v_{f}$ - 100)

$v_{f}$ = 106 m/s

5 0

2 years ago

Read 2 more answers

1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the

Juli2301 [7.4K]

The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

Time, t = 2s

Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

F = 6975N

Therefore, the force exerted on the car during this stop is 6975N

6 0

2 years ago

A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0Â° east of

koban [17]

<u>Answer:</u>

Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

Let the east point towards positive X-axis and north point towards positive Y-axis.

Speed of truck = 25 m/s north = 25 j m/s

Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

= (1.43 i + 1.00 j) m/s

Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

Magnitude of velocity = 26.04 m/s

Angle from positive horizontal axis = 86.85⁰

So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0

2 years ago

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

6 0

2 years ago