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Montano1993 [528]
2 years ago
11

A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p

oint A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.55 m up the incline from A, the block is moving up the incline at a speed of 7.50 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is \mu_k = 0.50. The mass of the spring is negligible.
Calculate the amount of potential energy that was initially stored in the spring. Take free fall acceleration to be 9.80 m/s².
Physics
1 answer:
andreyandreev [35.5K]2 years ago
7 0

Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

Mass of block = 1.60 kg

Angle = 30.0°

Distance = 6.55 m

Speed = 7.50 m/s

Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

U_{A}+k_{A}+w_{A}=U_{B}+k_{B}

U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2

Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

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Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d
DedPeter [7]

Answer:

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

x-component = -1.50 m/s

y-component = 3.90 m/s

Explanation:

Relative velocity of a body A relative to another body B, Vab, is given as

Vab = Va - Vb

where

Va = Relative velocity of body A with respect to another third body or frame of reference C

Vb = Relative velocity of body B with respect to that same third body or frame of reference C.

So, relative velocity can be given further as

Vab = Vac - Vbc

Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.

Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form

What is Pauli's velocity relative to Daniel?

Vpd = Vp - Vd

(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)

Vpd = Vpn - Vdn

Vpn = -Vnp = -(1.50î) m/s

Vdn = -Vnd = -(3.90ĵ) m/s

Vpd = -1.50î - (-3.90ĵ)

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

Hope this Helps!!!!

5 0
2 years ago
The Kyoto protocol works by _______.
Lapatulllka [165]
The answer is voluntary involvement! :D Welcome
7 0
3 years ago
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An eagle flies from its perch in a tree to the ground to capture and eat its prey. describe its energy transfromation
ddd [48]
10% energy is transferred from the prey to eagle and 90% energy is lost in the enviorment .
7 0
2 years ago
When an object accelerates, what about its motion changes? Question 1 options: Speed must change, but not velocity. Both speed a
QveST [7]

Answer:

The velocity must change but not speed.

Explanation:

  • Velocity is defined as the displacement by time. Whereas speed is expressed as the distance between two successive positions of the body to the time interval it took to travel.

                <em>Velocity,        V = D / t        m/s</em>

<em>                  Speed,          s = d /t          m/s        </em>  

  • Velocity is a vector quantity that has a magnitude and direction.
  • The speed is a scalar quantity having only the magnitude.
  • At any instant of time, the magnitude of the velocity is always equal to the magnitude of the speed. The magnitude of velocity, |<em>v </em>| = magnitude of speed, |<em>v </em>|. The magnitude is always positive
  • The acceleration of a body is defined as the rate of change of velocity to time.

                               <em>   a = (v - u) / t      m/s²</em>

  • If a body is accelerating, It varies its velocity with respect to time.
  • In case of uniform circular motion, the speed remains constant, but the velocity changes continuously.

So, in the case of circular motion if an object accelerates, velocity must change but speed remains constant.

5 0
3 years ago
A china dish falls from a height of 1.5 m above the floor. Calculate its velocity just before it
alexandr402 [8]

Answer:

12m/s

Explanation:

v^2=u^2+2as

v=?

u=0 (the dish was stationary before it fell)

a=9.81 m/s^2 (acceleration due to gravity/freefall)

s=1.5m (the drop height)

So: v^2=0+2.9.81.1.5 = 144.35415

and therefore v=sqrt 144.35415

12x12=144 so I'd say v=12m/s

6 0
2 years ago
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