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7nadin3 [17]
4 years ago
11

A spring stores 10. joules of elastic potential

Physics
2 answers:
Mekhanik [1.2K]4 years ago
6 0
The answer would be . Since we are looking for the spring constant you would need to use the formula
PEs =  \frac{1}{2} k {x}^{2}
. Then you'd substitute, for PEs and x.
10j=1/2k(.20m^2).
Then solve. k=500n/m
serg [7]4 years ago
6 0
<h2>Answer:</h2>

The spring constant is 50 N/m.

<h3>Explanation:</h3>

Spring constant is the specific property of a spring which tells about the stiffness of a spring.

It is the amount of force which is used to displace a spring one meter from its equilibrium position.

Its unit is N/m.

The formula of potential energy in a spring is :

P.E = 1/2Kx2

Here,

P.E = 10 joules

x = 0.2m

K = P.E × 2/x2

K = 10 × 2/0.4

k = 50 N/m

Hence the spring constant is 50 N/m.

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We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one
Len [333]

Answer:

A=1

B=-2

Explanation:

Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.

From part B as attached, it shows that the right option is C which is

2A+3B=-4

Substituting B with 3A-5 then we form the second equation as shown

2A+3(3A-5)=-4

By simplifying the above equation, we obtain

2A+9A-15=-4

Re-arranging, then

11A=-4+15

Finally

11A=11

A=1

To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain

B=3(1)-5=-2

Therefore, required values are 1 and -2

3 0
3 years ago
A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the
crimeas [40]

Answer:

Explanation:

  • The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
  • where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
  • The potential energy of the system is given as ; PE = mgh
  • The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
  • The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
  • Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

7 0
3 years ago
Define uniform and non uniform
cricket20 [7]

Answer:

When a body moves along a straight line with uniform speed or steady speed is called Uniform motion. When a body moves along a straight line but with variable or change in speed is called non-uniform motion.Hope this answer helps.

7 0
3 years ago
Read 2 more answers
A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of
andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where :

n_1 is the index of refraction of the 1st medium

n_2 is the index of refraction of the 2nd medium

\theta_1 is the angle of incidence (the angle between the incident ray and the normal to the boundary)

\theta_2 is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

n_1=1.00 (index of refraction of air)

n_2=1.50 approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

sin \theta_2 =\frac{n_1}{n_2}sin \theta_1

And since n_2>n_1, we have

\frac{n_1}{n_2}

And so

\theta_2

Which means that

The angle of incidence is greater than the angle of refraction

6 0
3 years ago
Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

6 0
3 years ago
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