Please answer the question in the picture below! Need it urgently!!!

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction

Korvikt [17]

Answer:

The magnitude of the impulse delivered to the baseball is 7.0 Ns

Explanation:

Given;

mass of the foul ball, m = 0.14 kg

initial velocity, u = 40 m/s

final velocity, v = 30 m/s in perpendicular direction

Impulse is given as change in momentum;

initial momentum in horizontal direction, Pi = mu

Pi = 0.14 x 40 = 5.6 Ns

final momentum in perpendicular direction, Pf = mv

Pf = 0.14 x 30

Pf = 4.2 Ns

The resultant impulse is given by;

J² = 5.6² + 4.2²

J² = 49

J = √49

J = 7.0 Ns

Therefore, the magnitude of the impulse delivered to the baseball is 7.0 Ns

5 0

3 years ago

Describe the conditions necessary for equilibrium on a balance.

tia_tia [17]

Answer:

Explanation:

A body under concurrent forces (concurrent forces are forces with their line of action acting at a point) is said to be in equilibrium if the sum total of the forces acting on the body is zero.

For forces acting on a balance, the sum of upward forces and the downward forced acting on the balance must be equal for the balance to be in equilibrium.

Also the sum of clockwise moment must be equal to the sum of anticlockwise moment acting on the balance.

Moment is the product of force and the perpendicular distance from the force to the pivot.

7 0

4 years ago

In a perfectly inelastic one-dimensional collision between two objects, what condition alone is necessary so that the final kine

Crank

<span>The condition alone that is necessary so that the final kinetic energy of the system is zero after the collision is that the objects must have momenta with the same magnitude but opposite directions.</span>

5 0

3 years ago

Read 2 more answers

1.

Kruka [31]

Answer:

The weight of the object X is approximately 3.262 N (Acting downwards)

The weight of the object Y is approximately 8.733 N (Acting downwards)

Explanation:

The question can be answered based on the principle of equilibrium of forces

The given parameters are;

The weight of Z = 12 N (Acting downwards)

The weight of the pulleys = Negligible

From the diagram;

The tension in the in the string attached to object Z = The weight of object Z = 12 N

The tension in the in the string attached to object X = The weight of the object X

The tension in the in the string attached to object Y = The weight of the object Y

Given that the forces are in equilibrium, we have;

The sum of vertical forces acting at a point, $\Sigma F_y$ = 0

Therefore;

$T_{1y} + T_{2y} + T_{3y} = 0$

$T_{1y} = -( T_{2y} + T_{3y} )$

Where;

$T_{1y}$ = The weight of object Z = 12 N

$T_{1y}$ = 12 N

$T_{2y}$ = The vertical component of tension, T₂ = T₂ × sin(24°)

∴ $T_{2y}$ = T₂ × sin(156°)

Similarly;

$T_{3y}$ = T₃ × sin(50°)

From $T_{1y} = -( T_{2y} + T_{3y} )$ , and $T_{1y}$ = 12 N, we have;

12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)

Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0

Therefore at point B, we have;

T₁ₓ + T₂ₓ + T₃ₓ = 0

The tension force, T₁, only has a vertical component, therefore;

∴ T₁ₓ = 0

∴ T₂ₓ + T₃ₓ = 0

T₂ₓ = -T₃ₓ

T₂ₓ = T₂ × cos(156°)

T₃ₓ = T₃ × cos(50°)

From T₂ₓ = -T₃ₓ, we have;

T₂ × cos(156°) = - T₃ × cos(50°)...(2)

Making T₃ the subject of equation (1) and (2) gives;

Making T₃ the subject of equation in equation (1), we get;

12 = -(T₂ × sin(156°) + T₃ × sin(50°))

∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))

Making T₃ the subject of equation in equation (2), we get;

T₂ × cos(156°) = - T₃ × cos(50°)

∴ T₃ = T₂ × cos(156°)/(-cos(50°))

Equating both values of T₃ gives;

(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))

-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))

∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283

∴ T₂ ≈ -8.02 N

From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;

T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717

∴ T₃ ≈ -11.4 N

The weight of the object X = -T₂ × sin(156°)

∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N

The weight of the object X ≈ 3.262 N (Acting downwards)

The weight of the object Y = -(T₃ × sin(50°))

∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N

The weight of the object Y ≈ 8.733 N (Acting downwards)

4 0

3 years ago