The Moon s escape speed will be smaller than Earth's.
The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.
The formula of escape speed is
where
v is escape velocity
G is universal gravitational constant
M is mass of the body to be escaped from
r is distance from the center of the mass
we can say that,
Escape speed depends on the gravity of the object trying to hold the spacecraft from escaping.
we know that,
The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second.
since, v ∝ g
The Moon s escape speed will be smaller than Earth's.
Learn more about escape speed here:
<u>brainly.com/question/15318861</u>
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The answer is : Open circuit
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s
Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s
Answer:
150 inches (12.5 ft)
Explanation:
The work done to lift the 500 pound block 3 inches should be the same as that to lift the 10 pond object a given distance.
The following is the equation one needs to solve:
![10 \,lb\,* \,x\,=\,500\,lb\,*\,3\,in\\10 \,lb\,* \,x\,=\,1500\,lb\,in\\](https://tex.z-dn.net/?f=10%20%5C%2Clb%5C%2C%2A%20%5C%2Cx%5C%2C%3D%5C%2C500%5C%2Clb%5C%2C%2A%5C%2C3%5C%2Cin%5C%5C10%20%5C%2Clb%5C%2C%2A%20%5C%2Cx%5C%2C%3D%5C%2C1500%5C%2Clb%5C%2Cin%5C%5C)
therefore solving for the distance "x" gives as the answer (in inches):
![10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in](https://tex.z-dn.net/?f=10%20%5C%2Clb%5C%2C%2A%20%5C%2Cx%5C%2C%3D%5C%2C1500%5C%2Clb%5C%2Cin%5C%5Cx%5C%2C%3D%5C%2C%5Cfrac%7B1500%5C%2Clb%5C%2Cin%7D%7B10%5C%2Clb%7D%20%5C%5C%5C%5Cx%5C%2C%3D150%5C%2Cin)
which can also be given in feet as: 12.5 ft
Answer:The mass of ball B is 10 kg.
Explanation;
Mass of ball A = ![M_A=5 kg](https://tex.z-dn.net/?f=M_A%3D5%20kg)
Velocity of the ball A before collision:![U_A=20 m/s](https://tex.z-dn.net/?f=U_A%3D20%20m%2Fs)
Velocity of ball A after collision=![V_A=10 m/s](https://tex.z-dn.net/?f=V_A%3D10%20m%2Fs)
Mass of ball B= ![M_B](https://tex.z-dn.net/?f=M_B)
Velocity of the ball B before collision:![U_B=10 m/s](https://tex.z-dn.net/?f=U_B%3D10%20m%2Fs)
Velocity of ball B after collision=![V_B=15 m/s](https://tex.z-dn.net/?f=V_B%3D15%20m%2Fs)
![M_AV_A+M_BV_B=M_AU_A+M_BU_B](https://tex.z-dn.net/?f=M_AV_A%2BM_BV_B%3DM_AU_A%2BM_BU_B)
![5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s](https://tex.z-dn.net/?f=5%20kg%5Ctimes%2010%20m%2Fs%2BM_B%5Ctimes%2015%3D5%20kg%5Ctimes%2020m%2Fs%2BM_B%5Ctimes%2010m%2Fs)
![M_B=10kg](https://tex.z-dn.net/?f=M_B%3D10kg)
The mass of ball B is 10 kg.