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777dan777 [17]
2 years ago
11

A toy plane has a mass of 2.5 kg and is 18 m above the ground. It is moving at 4.5 m/s. How much mechanical energy does the toy

plane have? 416 J 441 J 466 J 491 J
Physics
2 answers:
kodGreya [7K]2 years ago
6 0

Answer:

466 J

Explanation:

I just took the quiz

sleet_krkn [62]2 years ago
4 0
Mechanical energy is the sum of kinetic energy and potential energy, or E=Ek+Ep. So Ek=(1/2)*m*v² where m is the mass of the object and v is it's velocity. Mp=m*g*h where m is the mass, g=9.81 m/s² and h is the height of the object. So after we input the numbers the total mechanical energy is
E=(1/2)*2.5*(4.5²) + 2.5*9.81*18 = 25.3125 J + 441.45 J = 466.7625 J. The correct answer is E= 466 J. 
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You hold a piece of wood in one hand and a piece of iron in the other. both pieces have the same volume, and you hold them fully
aivan3 [116]
You hold a piece of wood in one hand and a piece of iron in the other. Both pieces have the same volume, and you hold them fully under water at the same depth. At the moment you let go of them, which one experiences the greater buoyancy force?<span>
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5 0
2 years ago
A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards
Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

Y = 1.6 X - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y =- \frac{1.6}{2(-0.032)} = 25

and maximum height is

y = 1.6\times 25 - 0.032\times 25^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0
2 years ago
Read 2 more answers
Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed
makkiz [27]
Using the given formula with v0=56 ft/s and h=40 ft 
h = -16t2 + v0t  
40 = -16t2 + 56t 
16t2 - 56t + 40 = 0  
Solving the quadratic equation:  
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
 We have two possible solutions  
t1 = (56+24)/32 = 2.5 
t2 = (56-24)/32 = 1  
So initially the ball reach a height of 40 ft in 1 second.
3 0
2 years ago
Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur
d1i1m1o1n [39]

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

                              mV + MV = mu + MU

                               V(m+M) = mu

                                 V = mu/(m+M)

Substituting the values in the above equation,

                                V = 0.04 Kg x 300 m/s  / (0.04 Kg+ 0.5 Kg)

                                    = 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

8 0
2 years ago
A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga
andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

Cp-Cv= R

Cp=7/2 R

Now by putting the values

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

Process is adiabatic

\Delta S_{surr}=0

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

\Delta S_{univ}= 20.79 +0

\Delta S_{univ}= 20.79

3 0
2 years ago
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