Please help me with this for science

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

3 years ago

Science questions!! Please help!!

kondor19780726 [428]

Answer:

1. b

2. c

3. b

4. a

5. a

6. c

7. c

8. d

7 0

2 years ago

A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth

Alex73 [517]

Answer:

$F_{Earth}$ = 15.57 N

$F_{Moon}$ = 2.60 N

$F_{Uranus}$ = 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

$F_{Earth}$ =m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

$F_{Moon}$ =m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

$F_{Uranus}$ =m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0

2 years ago

Which of the following cars have the most kinetic energy

faltersainse [42]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>

$\huge\boxed{OptionA}$

mass = 4 kg

velocity = 5 m/s^2

<h3 /><h3>Orange truck:</h3>

Mass= 2kg

Velocity = 7m/s^2

mass = 6 kg

velocity = 4m/s^2

<h3 /><h3>Green Car:</h3>

Mass = 8 kg

Velocity = 3 m/s^2

<h2>_____________________________________</h2><h2>SOLUTION:</h2>

By the equation of kinetic energy,

K.E = $\frac{1}{2} mv^2$

Where,

K.E is kinetic energy

m is mass

v is velocity

<h2>_____________________________________</h2><h3>Kinetic energy of Blue car:</h3>

Directly substitute the variables in the equation,

K.E = $\frac{1}{2}x4x5^2$

Simplify the equation,

K.E = 50 J

<h2>_____________________________________</h2><h3>Kinetic Energy of Silver Car:</h3>

Directly substitude the variable in the equation,

K.E = $\frac{1}{2}x6x4^2$

Simplify the equation,

K.E = 48J

<h2>_____________________________________</h2><h3>Kinetic Energy of Green Car:</h3><h3 />

Substitute the variables in the equation,

K.E = $\frac{1}{2}x8x3^2$

Simplify the Equation,

K.E = 36J

<h2>_____________________________________</h2><h3>Kinetic Energy of Orange Truck:</h3><h3 />

Substitute the variable,

K.E = $\frac{1}{2}x 2x7^2$

Simplify the equation,

K.E = 49J

As you can see that the highest value of kinetic energy is of Blue SUV thus it will be out answer.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h3 /><h3 /><h3 /><h3 /><h2 /><h2 />

6 0

2 years ago

What is the difference between habitats and niches?

jeyben [28]

Habitats is a place were someone or something lives, and idk what a ninch is

5 0

3 years ago

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