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Korolek [52]
3 years ago
14

There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________

Physics
1 answer:
salantis [7]3 years ago
8 0

Answer:

The bell has a potential energy of 8550 [J]

Explanation:

Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.

E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J]

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A train travels 55 km south along a straight track in 34 minutes. What is the train's average velocity in kilometers per hour?​
AURORKA [14]
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The speed of a 2 kg mass on a spring is 5 m/s as it passes through its equilibrium position. What is its frequency if the amplit
Lena [83]

Solution :

Given :

Mass of the object attached to the spring, m = 2 kg

Velocity of the object as it moves, V = 5 m/s

Amplitude of the object as it swings, A = 2.5 m

We have to find the frequency of the object.

We known,

$V_{max} = \omega A = 2 \pi f A$

Therefore, $f= \frac{V_{max}}{2 \pi A}$

$f= \frac{5}{2 \times 3.14 \times 2.5}$

f   = 0.32 Hz

Therefore the frequency of the object is 0.32 hertz when the amplitude of the 2 kg mass is 2.5 m moving a speed of 5 m/s.

3 0
2 years ago
Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the
inna [77]

Answer:

Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

                                     F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

                                     F_net,x = 0

- The forces as follows:

                                     Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

                                     Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

                                    - W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

                                     Fn = W + Fh*sin(θ)

                                     Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

                                     Ff = u*Fn

                                     Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

                                     Fh*cos(θ) = u*(m*g + Fh*sin(θ))

                                     Fh*(cos(θ) - u*sin(θ)) = u*m*g

                                     Fh = u*m*g / (cos(θ) - u*sin(θ))

5 0
3 years ago
A boy pulls a sled 46 m across a flat field with a force of 120 N. The force acts on the sled through a rope at a 23° angle with
ivann1987 [24]

Answer:

5520

Explanation:

w=f×d

w=120×46

w=5.52×10^3

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