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Gnom [1K]
3 years ago
6

A museum curator moves artifacts into place on various different display surfaces. If the curator moves a 145 kg aluminum sculpt

ure across a horizontal steel platform with a force of 668N, what is the coefficient of kinetic friction?
Physics
2 answers:
NikAS [45]3 years ago
8 0
We are given the mass of an <span>aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47</span>
shepuryov [24]3 years ago
8 0

Answer:

\mu_k = 0.47

Explanation:

As we know that the mass of the curator is

M = 145 kg

now the Force of kinetic friction on the curator is given as

F_k = \mu_k F_n

now we know that the normal force on the curator is counter balanced by the weight of the curator

so we can say

F_n = mg

F_n = (145)(9.8)

F_n = 1421 N

Now from above formula we have

668 = \mu_k (1421)

\mu_k = \frac{668}{1421}

\mu_k = 0.47

You might be interested in
in a car moving at constant acceleration, you travel 230 mm between the instants at which the speedometer reads 40 km/hkm/h and
Ronch [10]

The acceleration of the car is 0.8049m/s^{2}.It takes 13.802s to travel the 230 m.

<h3>What is acceleration?</h3>

In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it. Newton's Second Law states that the combined effect of two factors determines how much an item accelerates: 

(i) It follows that  the magnitude of the net balance of all external forces acting on the object is directly proportional to the magnitude of this net resulting force, and

(ii) the mass of the thing, depending on the materials out of which it is constructed, is inversely proportional to the mass of the thing.

Calculations:

40 km/hr ----- 11.11m/s

80 km/hr ----- 22.22m/s

v^{2} -u^{2} =2as\\22.22^{2} - 11.11^{2} = 2 x a x 230\\ 370.296=460a\\ a= 0.8049m/s^{2} \\

Time taken

v-u=at

22.22-11.11= 0.8049 x t

t=13.802s

To learn more about acceleration ,visit:

brainly.com/question/2303856

#SPJ4

4 0
2 years ago
Find the value of 15.0 N in pounds. Use the conversions 1slug=14.59kg and 1ft=0.3048m. Express your answer in pounds to three si
kotykmax [81]

Given the equivalent value to convert the units from kilograms to slug and feet to meters, we will proceed to define the equivalence of the 'Newton' in simplified units, that is,

1N=1kg\cdot m\cdot s^{-2}

Then,

15.0N=15.0kg\cdot m\cdot s^{-2}

Converting this value to British units we have that

15N = 15.0kg\cdot m\cdot s^{-2} (\frac{1slug}{14.59kg})(\frac{1ft}{0.3048m})

15N = 3.37 slug \cdot ft \cdot s^{-2} (\frac{1 lb \cdot ft^{-1}}{1slug})

15N =3.37lb

Therefore the value of 15.0 N in pounds is 3.37 lb.

6 0
3 years ago
2. In the same tournament, a player is positioned 35 m (40° W of S] of the net. He shoots the puck
andrezito [222]

Answer:

The displacement of the net from player 2 in component form = (-47.498î - 26.812j)

The displacement of the net from player 2 in statement form is 54.54 m and 29.44° (S of W) or 60.56° (W of S)

Explanation:

The sketch of the bearings described in the question is presented in the attached image to this solution.

Method 1

Using component method

Taking the player 1's position as the origin,

The displacement of the player 2 from the origin is (25î) m

The displacement of the net from the origin is 35[(sin θ)î + (cos θ)j]

But θ is the angle of the net's displacement reading from the positive x-axis in the anticlockwise direction. θ = 230°

Displacement of the net from the origin = 35[(cos 230°) + (cos 230°)]

= 35[-0.6428î - 0.7660j]

= (-22.498î - 26.812j) m

In component form, taking note of the directions of the respective displacements calculated (check the attached image)

(The displacement of the net from player 1) = (The displacement of player 2 from player 1) + (The displacement of the net from player 2)

Since we have agreed that player 1 is the origin

(The displacement of the net from origin) = (The displacement of player 2 from origin) + (The displacement of the net from player 2)

(-22.498î - 26.812j) = (25î) + (The displacement of the net from player 2)

The displacement of the net from player 2 = (-22.498î - 26.812j) - (25î) = (-47.498î - 26.812j)

The magnitude of this displacement = √[(-47.498)² + (-26.812)²]

= √(2,256.060004 + 718.883344) = 54.54 m

Direction = tan⁻¹ (-26.812/-47.498) = 209.44° (the signs on the components show that the direction is the third quadrant from the positive x-axis in the anti-clockwise direction)

Hence, the displacement of the net from player 2 is 54.54 m and 29.44° (S of W)

Method 2

Using trignometry,

We will use cosine and sine rule to obtain the required magnitude and direction of the displacement of the net from player 2

Cosine rule

Magnitude = √[35² + 25² - (2×25×35×cos 130°)] = √2,974.8783169514 = 54.54 m

Sine rule

(Sin θ)/35 = (Sin 130°)/54.54

Sin θ = (35 × Sin 130°)/54.54 = 0.4916

θ = Sin⁻¹ (0.4916) = 29.44°

This answer matches the answers from method 1.

Hope this Helps!!!

5 0
3 years ago
At the presentation ceremony, a championship bowler is presented a 1.60-kg trophy which he holds at arm's length, a distance of
nikdorinn [45]

Answer:

a)10.28 Nm

b)9.93 Nm

Explanation:

Let g = 9.81m/s2. First we can calculate the weight of the trophy

W = mg = 1.6 * 9.81 = 15.696 N

(a) The torque is product of force and its moment arm

T = WL = 15.696 * 0.655 = 10.28 Nm

(b) Suppose his arm makes an angle of 15 degree with respect to the horizontal line. We can still calculate the arm length, or the horizontal distance from the trophy to the champion:

L_2 = Lcos(15^0) = 0.655cos(15^0) = 0.655*0.966 = 0.633 m

Again, torque is product of force and its moment arm

T_2 = WL_2 = 15.696 * 0.633 = 9.93 Nm

8 0
4 years ago
The term atomic number most nearly means
dimulka [17.4K]
D. is the answer to your question
5 0
3 years ago
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