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3 years ago

13

An athlete stretches a spring an extra 28.6 cm beyond its initial length. how much energy has he transferred to the spring, if t

he spring constant is 52.9 n/cm?

1 answer:

g100num [7]3 years ago

6 0

PE = 0.5 × k × x²

PE potential Energy
k spring constant
x stretch/compression of the spring

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If the Net Force of the object is 30 N to the left, and mass is 3 kg what is the objects acceleration?

kaheart [24]

Answer:

Oops i put the answer in a comment,

-10 to the left m/s^2

Explanation:

7 0

3 years ago

Lions can run at speeds up to approximately 80.0 km / h. A hungry 109 kg lion running northward at top speed attacks and holds o

sukhopar [10]

Answer:

17.34 m/s

Explanation:

Given:

Mass of lion (m₁) = 109 kg

Initial speed of lion (v₁) = 80.0 km/h (Northward direction)

Mass of gazelle (m₂) = 39.0 kg

Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)

Final velocity of both lion and gazelle is, $v_f=?$

First, let us convert the speeds from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore,

$v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s$

Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.

Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.

So, initial momentum is given as:

$P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle$

Now, we calculate P₁ and P₂.

$P_1=m_1v_1=(109\ kg)(22.22\ m/s) = 2421.98\ Ns\\\\P_2=m_2v_2=(39\ kg)(21.81\ m/s) = 850.59\ Ns$

Therefore, the net initial momentum of the system is given as:

$P_i=\sqrt{(2421.98)^2+(850.59)^2}=2567\ Ns$

The final momentum of the system is given as:

$P_f=(m_1+m_2)(v_f)\\\\P_f=(109+39)v_f\\\\P_f=148v_f$

From the law of conservation of momentum, the final momentum is equal to the initial momentum. So,

$P_f=P_i\\\\148v_f=2567\\\\v_f=\frac{2567}{148}=17.34\ m/s$

Therefore, the final speed of the lion-gazelle system is 17.34 m/s

3 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago

Now suppose you carry out a second Thomson experiment with a different beam that contains two types of particles. In particular,

Tems11 [23]

Answer:

Two off-centered spots in the first phase of the experiment; one centered spot in the second phase of the experiment.

Explanation:

If two particles are selected in which both have the same electron mass and the same velocity, but one of the particles has a charge and the other particle has a charge of 2e. During the first stage of the experiment, the two particles have an electric force equal to F = Eq in the entire vertical direction. The acceleration of particle is equal to a = (Eq)/m.

In the second part of the experiment, the magnetic field cancels the electric field. In this way, the electric force and the magnetic force cancel each other out. Therefore, the net force acting on each particle is equal to zero.

Because these two forces cancel each other out, the particles fail to create two off-center points on the screen in the second part of the experiment. Also, if the loads are different, the deviation is also different. In this way, an off-center point cannot be achieved in the first part of the experiment. There will be two off-center points.

8 0

3 years ago

A wire with a resistance of 20 is connected to a 12 battery. What is the current flowing through the wire?

Dima020 [189]

15.49 should be the answer if that is 12 watt battery.

3 0

3 years ago