Is there more forms of energy of there than :

A 2.85 kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 875 N/m. The spring

Westkost [7]

Answer:

A) V = 0.324 m/s

B) V = 0.56 m/s

Explanation:

A) Change in gravitational potential energy = mass * gravity * height change

Change in G.P.E = 2.85 * 9.81 * 0.058 = 1.6215 J

Energy transferred to spring = 0.5 * k * x^2

Here k = 875 N/m

And x = 0.058 m

Thus energy transferred to spring = 1.47175 J

The G.P.E is converted to spring potential energy and kinetic energy as it moves down. So the difference in energy is accounted by kinetic energy as follows:

Kinetic energy = G.P.E - Spring P.E

Kinetic energy = 1.6215 - 1.47175 = 0.14975 J

We can now find the speed using this kinetic energy:

Kinetic energy = 0.14975

0.5 * mass * velocity^2 = 0.14975

0.5 * 2.85 * V^2 = 0.14975

V = 0.324 m/s

B) The fish accelerates because the force on it are unbalanced. These forces are the weight of the fish, and the force of the spring stopping it.

As long as the weight of the fish is more than the upward force of the spring, the fish will continue to accelerate. Using this knowledge, we can deduce that the speed is maximum when the weight and spring force are equal. Thus we set them equal and find out the displacement first:

Weight = spring force

2.85 * 9.81 = 875 * Displacement

Displacement = 0.03195 m

Similarly as (A):

Change in G.P.E = 2.85 * 9.81 * 0.03195 = 0.8933 J

Spring P.E = 0.5 * 875 * 0.03195^2 = 0.4466 J

Kinetic energy = 0.8933 - 0.4466 = 0.4467 J

0.5 * mass * V^2 = 0.4467

0.5 * 2.85 * V^2 = 0.4467

V = 0.56 m/s

4 0

3 years ago

A bowling ball has a mass of 5 kg what happens to its momentum when its speed increases from 1m/s to 2m/s?

Mama L [17]

Here, Initial momentum = mu = 5*1 = 5 Kg m/s
Final momentum = mv = 5*2 = 10 Kg m/s

So, Momentum has been increased from 5 Kg m/s to 10 Kg m/s. Hence, Your Final answer is option B

Hope this helps!

8 0

3 years ago

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W = 0

N = W

N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax , ax=0

282 N - 309.58 N = 90*a

a= (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

8 0

3 years ago

Why solar panels are usually painted black

spayn [35]

Answer:

The silicon used to make monocrystalline solar cells has a high level of purity. ... Because of the way light interacts with a monocrystalline silicon layer, monocrystalline solar panels appear black in color.

3 0

3 years ago

Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to

Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

I will take two assumptions:

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the linear density of the stretched wire) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new charge must be 10 times the original charge, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0

3 years ago