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Lesechka [4]
3 years ago
11

The drawing represents a mountain bike trail. A rider began at Point 1 on the trail and stopped at Point 3. At which point did h

e have the least potential energy?

Physics
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer:

Point 2.

Explanation:

Potential energy is simply defined as the energy stored in an object due to its position. It is can be represented mathematically by:

P.E = mgh

Where:

P.E is the potential energy.

m is the mass of the object.

g is acceleration due to gravity.

h is the height to which the object is located.

From the above equation, we can thus say that potential energy depends on the height of the object since the mass of the object is always constant i.e as the height of the object increase, the potential energy also increases and as the height of the object decrease, the potential energy also decreases.

Now, considering the diagram in the question given, we can see that point 2 is the lowest height to which the rider is located. At this point i.e point 2, the rider will have the least potential energy.

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A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig
Blababa [14]

Answer:

<u>Please Mark As Brainliest!!</u>

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a)  If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf   ⇒  I₁ * ω₁  = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)  

W = 7.03 J

4 0
3 years ago
A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position
kaheart [24]

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

         we know that,

                      Work = (1/2) k x²

                          704 = (1/2) × 500 × x²

                            x = 1.67 m

A spring stretched for 1.67 m distance.

4 0
3 years ago
If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
solong [7]
Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s

3 0
3 years ago
How can two strengths be added together
Ket [755]
By using mind and dna
6 0
3 years ago
A 1.15 kg book is at rest on the table. What is the magnitude of the normal force that the table is exerting on the book?
Agata [3.3K]

Answer:

11.27N

Explanation:

Given parameters:

Mass of the book  = 1.15kg

Unknown:

Magnitude of the normal force  = ?

Solution:

The normal force is the vertical force exerted by a body on an object.

It can be described as the weight of an object.

 Normal force  = mass x acceleration due to gravity

 Normal force  = 1.15 x 9.8  = 11.27N

5 0
3 years ago
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