Ask question

Ask question

All categories

3 years ago

8

5) A girl with 40kg mass is riding an 8 kg bicycle. At the top of the hill her speed is 6.0 m/s, and her speed triples at the bo

ttom of the hill. The hill is 15.0 m high. What is the work done by the non conservative forces?

1 answer:

Sedaia [141]3 years ago

6 0

Answer:

Work = 6912 joules

Explanation:

Non-conservative forces are dissipative forces such as friction or air resistance. These forces take energy away from the system as the system progresses, energy that you can't get back. These forces are path dependent; therefore it matters where the object starts and stops.

Total mass = 40 + 8 = 48kg

Initial speed u= 6 m/s

Final speed v = 3*initial

Final speed v = 3* 6 = 18 m/s

Distance s = 15

Acceleration a is?

V² = U² + 2aS

18² = 6² + 2a*15

324 = 36 + 30a

324-36= 30a

288 = 30a

288/30 = a

9.6= a

Force = mass* acceleration

Force = 48*9.6

Force = 460.8N

Work = force*distance

Work = 460.8*15

Work = 6912 joules

You might be interested in

What type of electromagnetic wave does a light bulb and a radio antenna release?

andrew-mc [135]

I believe that a light bulb releases visible light and a radio antenna releases a radio waves

4 0

3 years ago

Do quasars reside within or without side of galaxies?

sveticcg [70]

They almost entirely reside within galaxies because quasars are a subset of blackholes with a large and fast enough accretion disk to generate a beam of interstellar material perpendicular to itself. This typically only occurs in the largest black holes at the center of galaxies (supermassive blackholes) or at least stellar black holes---which still occur within galaxies because the material is necessary to form them.

6 0

3 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

3 0

3 years ago

Enter an expression for the force constant for the floating raft, in terms of L, g, and the density of water, ρ.

andriy [413]

Answer:

K = ρL²g

Explanation:

Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;

Then:

F = upthrust ( restoring force) = weight of the liquid displaced.

$F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g$

where;

A = L²

$\rho_{\omega} = \rho$

F = ky.

Then,

$Ay \rho g = ky$

$L^2y \rho g = ky$

Divide both sides by y

K = ρL²g

3 0

3 years ago

A trio of students push a 65 kg crate. The first student pushes 31 N [e], the second student pushes 28 N [s] and the third stude

Verdich [7]

Answer:

The free body diagram is attached.

Explanation:

A force of 31[N] to the east, the second force goes to the south and it is equal to 28[N], the third force goes to the west and it is equal to 39 [N].

We can consider the crate as a particle. And all the forces are acting over the particle.

5 0

3 years ago