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2 years ago

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a mass of 10kg is placed on a horizontal table with coefficient of friction is 0.5. if the mass is static, determine weight of t

he object against the horizontal table and the limiting frictional force.

1 answer:

zloy xaker [14]2 years ago

5 0

Answer:

10

Explanation:

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The more twists per foot in a pair of wires, the more resistant the pair will be to ____.

Murljashka [212]

The more twist per foot in a pair of wires, the more resistant the pair will be to cross talk. A cross talk in network planning and design is a disturbance produced by electromagnetic interference beside a circuit or a cable pair. A telecommunication signal interrupts a signal in an adjacent circuit and can source the signals to turn out to be confused and cross over each other.

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3 years ago

What is the mass of an object that has a weight of 80.0 N?

Ira Lisetskai [31]

Explanation:

SUPONIENDO QUE LA ACELERACIÓN DE LA GRAVEDAD ES $9.80 m/s^{2}$

USANDO LA SEGUNDA LEY DE NEWTON:

<em>m</em> = 80.0 N/ $9.80 m/s^{2}$ = 8.16 kg

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3 years ago

Read 2 more answers

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

x = ( 0 - 15²) / 2 (-3.14)

x= 35.8 m

The shortest braking distance is 35.8 m

7 0

2 years ago

A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she

klasskru [66]

Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0

2 years ago

Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the

makvit [3.9K]

Answer:

$\mu_k=0.101$

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

$\dfrac{1}{2}kx^2=\mu_k mgh$

$\mu_k=\dfrac{kx^2}{2mgh}$

$\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}$

$\mu_k=0.101$

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0

2 years ago