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Svetradugi [14.3K]
3 years ago
15

A wheel with a weight of 388 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it i

s rotating at an angular velocity of 23.6 rad/s . The radius of the wheel is 0.591 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3482 J .
Calculate h. Use 9.81m/s2 for the acceleration due to gravity.
Physics
1 answer:
Alisiya [41]3 years ago
4 0

the correct answer is 14m

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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i
drek231 [11]

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = \frac{28}{10} × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

8 0
3 years ago
A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizont
PIT_PIT [208]

Answer:

v' = 2.83 m/s

Explanation:

Velocity of wave in stretched string is given by the formula

v = \sqrt{\frac{T}{\mu}}

here we know that

T = 4 N

also we know that linear mass density is given as

\mu = 1 kg/m

so we have

v = \sqrt{\frac{4}{1}} = 2 m/s

now the tension in the string is double

so the velocity is given as

v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s

v' = 2.83 m/s

4 0
3 years ago
A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.
Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

a=\frac{v^{2} }{r}, where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

a=\frac{3.5^{2} }{8.5} \\a=1.4

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0
3 years ago
16. For this table of data, how should the y-axis be labeled (with units)?
vampirchik [111]

Answer:

The y-axis should be labelled as W in Newtons (kg·m/s²)

Explanation:

The given data is presented here as follows;

Mass (kg)            {}        Newtons (kg·m/s²)

3.2                      {}           31.381

4.6             {}                    45.1111

6.1              {}                    59.821

7.4              {}                    72.569

9                {}                     89.241

10.4              {}                   101.989

10.9              {}                  106.892

From the table, it can be seen that there is a nearly linear relationship between the  amount of Newtons and the  mass, as the slope of the data has a relatively constant slope

Therefore, the data can be said to be a function of Weight in Newtons to the mass in kilograms such that the weight depends on the mass as follows;

W(m) in Newtons = Mass, m in kg × g

Where;

g is the constant of proportionality

Therefore, the y-axis component which is the dependent variable is the function, W(m) = Weight of the body while the x-axis component which is the independent variable is the mass. m

The graph of the data is created with Microsoft Excel give the slope which is the constant of proportionality, g = 9.8379, which is the acceleration due to gravity g ≈ 9.8 m/s²

We therefore label the y-axis as W in Newtons (kg·m/s²)

6 0
3 years ago
2. A company hires a security firm to patrol their stores and watch out for thieves. What is this an example of?
Lunna [17]

Answer:

someone hiring someone

Explanation:

4 0
3 years ago
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