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Vikki [24]
2 years ago
8

A rubber band has potential energy of 5 J. If the spring constant of the rubber band is 50 N/m, what is the displacement of the

rubber band?
Physics
2 answers:
Valentin [98]2 years ago
7 0
To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:

PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m

Therefore, the displacement would be 0.32 m.
Allisa [31]2 years ago
7 0
The correct answer is 0.4 m on Edge.
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A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
Pie

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

7 0
2 years ago
Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster
Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g

The acceleration is 1.10261 times g

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s

In mph

186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph

The speed of the dragster is 416.17506 mph

5 0
2 years ago
Which of the following strategies can help Earth's coal supply last longer?
ivanzaharov [21]
D. Teach the public energy conservation
7 0
2 years ago
Read 2 more answers
A student decides to give his bicycle a tune up. He flips it upside down (so there’s no friction with the ground) and applies a
Alinara [238K]

Answer:

Tangential velocity = 10.9 m/S

Explanation:

As per the data given in the question,

Force = 20 N

Time = 1.2 S

Length = 16.5 cm

Radius = 33.0 cm

Moment of inertia = 1200 kg.cm^2 = 1200 × 10^(-4) kg.m^2

= 1200 × 10^(-2) m^2

Revolution of the pedal ÷ revolution of wheel = 1

Torque on the pedal = Force × Length

= 20 × 16.5 10^(-2)

= 3.30 N m

So, Angular acceleration = Torque ÷ Moment of inertia

= 3.30 ÷ 12 × 10^(-2)

= 27.50 rad ÷ S^2

Since wheel started rotating from rest, so initial angular velocity = 0 rad/S

Now, Angular velocity = Initial angular velocity + Angular Acceleration × Time

= 0 + 27.50 × 1.2

= 33 rad/S

Hence, Tangential velocity = Angular velocity × Radius

= 33 × 33 × 10^(-2)

= 10.9 m/S

7 0
2 years ago
Which of the following statements are true about an object in two-dimensional projectile motion with no air resistance? (There c
ki77a [65]

Answer:

The correct answers are

The following statements are true about an object in two-dimensional projectile motion with no air resistance

D) The speed of the object is zero at its highest point.

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

Explanation:

A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

C) The acceleration of the object is zero at its highest point.

False, the acceleration is constant in magnitude throughout the motion

D) The speed of the object is zero at its highest point.

True, the direction of motion changes at the highest point from hence the body comes to rest and the speed is zero

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

6 0
2 years ago
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