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aleksandrvk [35]
3 years ago
10

Which acceleration does the force of earth’s gravity produce?

Physics
1 answer:
aev [14]3 years ago
5 0

I think its A

hope this helps

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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
If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.
sashaice [31]
I just had this question, the awnser is A.

6 0
3 years ago
Read 2 more answers
A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the
olasank [31]

Explanation:

The given data is as follows.

   Spring constant (k) = 78 N/m,     \theta = 30^{o}

 Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

                mgh sin \theta = \frac{1}{2}kx^{2}

       h = \frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}

          = \frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}

          = 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0
3 years ago
A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi
Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0
3 years ago
How many days are in 345000 minutes? (Set up by dimensional analysis)​
Komok [63]
239.583 days is the answer
3 0
2 years ago
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