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Pavel [41]
2 years ago
14

Ashkon throws a basketball across the court to his teammate. The ball has 57 J of potential energy and 61 J of kinetic energy. W

hat is the total mechanical energy of the basketball?
Physics
2 answers:
Gwar [14]2 years ago
8 0
The total mechanical energy of the ball is the sum of its potential energy U and its kinetic energy K, therefore:
E=U+K=57 J+61 J=118 J
so, the total mechanical energy of the basketball is 118 J.
s2008m [1.1K]2 years ago
7 0

The total mechanical energy of the ball is the sum of its potential energy U and its kinetic energy K, therefore:

E=U+K=57 J+61 J=118 J

so, the total mechanical energy of the basketball is 118 J.

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The difference between the two molar specific heats of a gas is 8000J/kgK. If the ratio of the two specific heats is 1.65, calcu
Serjik [45]

Answer:

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Explanation:

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4 0
2 years ago
Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two op
Nostrana [21]

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = v_{ex}.ln(\frac{m_{f}}{m_{e}} )

Where:

v_{ex} is relationship between exhaust velocity and specific impulse

\frac{m_{f}}{m_{e}} is the porpellant fraction, also written as MR.

The relationship v_{ex} is: v_{ex} = g_{0}.Isp

To determine the fraction:

Δv = v_{ex}.ln(\frac{m_{f}}{m_{e}} )

ln(MR) = \frac{v}{v_{ex}}

Knowing that change in velocity is Δv = 9.6km/s and g_{0} = 9.81m/s²

<u>Note:</u> Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

<u />

<u>Part 1</u>: Isp = 450s

ln(MR) = \frac{v}{v_{ex}}

ln(MR) = \frac{9.6.10^{3}}{9.81.450}

ln (MR) = 2.17

MR = e^{2.17}

MR = 8.76

<u>Part 2:</u> Isp = 2000s

ln(MR) = \frac{v}{v_{ex}}

ln (MR) = \frac{9.6.10^{3}}{9.81.2.10^{3}}

ln (MR) = 0.49

MR = e^{0.49}

MR = 1.63

8 0
2 years ago
When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant
weqwewe [10]

Answer:

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

       Q = A v

The area of ​​a circle is

      A = π r2 = π d²/4

     

     v = Q / A = Q 4 / pi d²

     v = 0.006 4/π 0.08²

     v =  1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

     P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0  

Rho (water) = 1000 kg / m³

      P₁ + rho y₁ = P₂ + ½ rho v₂²

      P₁ = P₂ + ½ rho v₂² - rho g y₁

      P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 10⁵ +708.5  - 9800

      P₁ =  92208.5Pa

      P₁ = 0.922 10⁵ Pa

8 0
2 years ago
An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

3 0
2 years ago
The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet
Naily [24]

Answer:

Speed of comet before collision is

v_{2_{i}}=-2.5\times10^{3}\quad km/h

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h

Case is considered as partially inelastic collision, by conservation of momentum

m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h

8 0
2 years ago
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