How many windows should the real tree house have

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Brut [27]

Answer:

The gravity does a work of - 117.6 Joules.

The tension does not do work as the force is perpendicular to the direction of motion at any point in the trajectory.

Explanation:

The work done by the gravity simply is the difference in gravitational potential energy multiplied by -1:

$W_g = - \Delta E_p = - (mgh_f - m g h_i)$

where m is the mass of the ball, g is the acceleration due to gravity, $h_f$ is the final height and $h_i$ is the initial height.

So, if the radius is 2.00 m, then the difference of height will be 4 meters:

$W_g = - mg (h_f - h_i)$

$W_g = - 3.00 \ kg \ 9.8 \frac{m}{s^2} \ 4 \m$

$W_g = - 117.6 Joules$

As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:

$dW = \vec{F} d\vec{r} = 0$

6 0

2 years ago

A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally

leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration a of

a = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ F = n - mg = 0

where ∑ F is the net force, n is the magnitude of the normal force, and mg is the weight of the block - it follows that n = mg ; and

• the net horizontal force is

∑ F = F - f = ma

where F is the applied force, f is kinetic friction, m is the block's mass, and a is the acceleration found earlier. F stretches the spring by x = 0.250 m, so we have

F - f = kx - µn = kx - µmg = ma

where k is the spring constant and µ is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ F = n - mg = 0

so that n = mg again; and

• the net horizontal force is

∑ F = F - f = 0

This time, F stretches the spring by y = 0.0544 m, so we have

F - f = ky - µmg = 0

Solve the equations in boldface for k and µ :

kx - µmg = ma

ky - µmg = 0

==> k (x - y) = ma

==> k = ma / (x - y)

==> k = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

ky - µmg = 0

==> µ = ky / (mg)

==> µ = (182 N/m) (0.0544 m) / ((17.6 kg) g) ≈ 0.0574

3 0

3 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

Hooke's law. The distance d when a spring is stretched by a hanging objective varies directly as the weight w of the object. If

Fofino [41]

Hooke's law is F=-kx, which means the elastic force contained by the spring is a product of the distance it stretches and its spring constant, but the direction of the force is opposite that of the displacement. We calculate as follows:

(3 kg)(9.8 m/s^2) = -k(-0.38 m)
k = 77.4

Then use k to find the new displacement, again using Hooke's law:
(7 kg)(9.8 m/s^2) = -(77.4)x
x = -0.89 m

7 0

3 years ago

Which of the following are examples of projectile motion?

denis-greek [22]

Answer:

A or C I would pick C

Explanation:

8 0

3 years ago

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