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3 years ago

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A 1.5 kg bird is gliding at a height of 12 m with a speed of 3.8m/s. What is the kinetic energy of the bird, to the nearest joul

e?

1 answer:

Georgia [21]3 years ago

5 0

A 1.5 kg bird is gliding at a height of 12 m with a speed of 3.8m/s. The kinetic energy of the bird is 10.83 joules.

Explanation:

Kinetic energy can be defined as,The kinetic energy (KE) of an object is the energy that the object possesses due to its motion.

The Kinetic energy can be calculated by using formula,

Kinetic Energy: KE = 1/2 (mv 2)

Where, m = Mass, v = Velocity.

Here in this case the bird mass is 1.5kg and is gliding with velocity 3.8m/s

hence, KE= 1/2*(1.5)×(3.8)^2

=0.5×1.5×3.8×3.8

=10.83Joules

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A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is

Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

P₁ V₁ = P₂ V₂

P_atm V₁ = (P_atm + ρ g H) V₂

V₂ = V₁ P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

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3 years ago

When looking at a food label, to be considered a good source of a nutrient, the nutrient must be? A.More than 20% of your daily

aleksklad [387]

Answer:

A. More than 20% of your daily recommended amount.

Explanation:

Reading food labels can be tricky. The percent daily value listed on the right of all food labels lets you know what percent out of the recommended daily intake of each nutrient you are consuming in that specific food.

To check if the food you're consuming is a good source of that nutrient you need in higher amount, the nutrient must be labeled 20% or higher.

The rule used here is called the 5/20 rule. According to this rule, A nutrient that is 5% or below is considered less and a nutrient which is labeled 20% or higher is considered good enough in that food source.

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3 years ago

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

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Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

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2 years ago

Which type of energy is thermal energy a form of?

mr Goodwill [35]

Answer:

Kinetic Energy

Explanation:

Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.

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3 years ago

Read 2 more answers

A 4-L pressurecooker has an operating pressure of175 kPa. Initially, one-half of the volume is filled with liquid and the other

Soloha48 [4]

To solve this problem, it is necessary to apply the concepts related to the Energy balance and the mass balance that allow us to find in each state the data necessary to find the total Power of the system through heat exchange.

From the tables of properties of the Water it is possible to obtain at the given pressures the values of the specific volume, the specific energy and the specific enthalpy.

Given these pressures then we have to

$P_1 = 174kPA$

$\Rightarrow v_f = 0.001057m^3/kg\\\Rightarrow v_g = 1.0037m^3/kg\\\Rightarrow u_f = 486.82kJ/kg\\\Rightarrow u_g = 2524.5kJ/kg$

$P_2 = 175kPa \rightarrow$ Saturated vapor

$\Rightarrow v_2 = v_g = 1.0036 m^3/kg\\\Rightarrow v_2 = u_g = 2524.5kJ/kg$

$P_e = 175kPa \rightarrow$ Saturated vapor

$V = 4L$

Considering the process performed, the kinetic and potential energy can be neglected as well as the work involved by specific interactions in the system. Although it is an unstable process it can be treated as a uniform flow process. Considering these expressions we can perform a mass balance for which

$m_{in}-m_{out} = \Delta m_{system} \rightarrow m_e = m_1-m_2$

Similarly through the energy balance you can get that

$E_{in}-E_{out} = \Delta E_{system}$

$Q_{in} -m_eh_e = m_2u_2-m_1u_1 \Rightarrow$ Since $W=ke=pe=0$

The initial mass, initial internal energy, and final mass in the tank are

$m_1 = mf_+m_g = \frac{V_f}{v_f}+\frac{V_g}{v_g}$

$m_1 = \frac{0.002m^3}{0.001057m^3/kg}+\frac{0.002m^3}{1.0036m^3/kg}$

$m_1 = 1.893+0.002= 1.895Kg$

At the same time the internal energy can be defined from the mass in state 1 as,

$U_1 = m_1 u_1$

$U_1 = m_fu_f+m_gu_g$

$U_1 = 1.893*486.82+0.002*2524.5$

$U_1 = 926.6kJ$

The calculation of mass in state 2 can be defined as

$m_2 = \frac{V}{v_2} = \frac{0.004m^3}{1.0037m^3/kg}$

$m_2 = 0.004Kg$

Then from the mass and energy balances,

$m_e = m_1-m_2\\m_e = 1.895-0.004\\m_e = 1.891Kg$

In this way the calculation of the heat of entry would be subject to

$Q_{in} = m_eh_e+m_2u_2-m_1u_1$

$Q_{in} = 1.891*2700.2+0.004*2524.5-926.6$

$Q_{in} = 4188kJ$

Therefore the Power would be given as

$\dot{Q} = \frac{Q}{\Delta t} = \frac{4188kJ}{3600s} = 1.163kW$

Therefore the highest rate of heat transfer allowed is 1.163kW

3 0

3 years ago