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Novay_Z [31]
3 years ago
10

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is _______.a) g/. b) g/4. c) 2g. d) g/2. e) g.
Physics
1 answer:
Liula [17]3 years ago
4 0

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

      v = v₀ - a t

     a = (v₀- v) / T

On planet X

    v = v₀ - a' t’

    a ’= (v₀-v) / 2T

Let's substitute the land values ​​in plot X

     a’= a / 2

Now let's use Newton's second law

       W = ma

      m g = m a

      a = g

We substitute

      a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

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Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa
Lorico [155]

Answer:r_2=11.81 cm

Explanation:

Given

m_1=2.2\times 10^{-8} kg

m_2=4.8\times 10^{-8} kg

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

qV=\frac{mv^2}{2}

v=\sqrt{\frac{2qV}{m}}

and we know

Force due to magnetic field will Provide centripetal Force

qvB=\frac{mv^2}{r}

B=\frac{\sqrt{\frac{2Vm}{q}}}{r}

and B is equal for both particles

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r_2^2=\frac{4.8}{2.2}\times 8^2

r_2=11.81 cm

4 0
3 years ago
What is the difference between direct current and alternating current?
IRINA_888 [86]
In a direct current, the electric charge, or current, only flows in one direction. In an alternating current, the electric charge changes periodically.  
5 0
4 years ago
Read 2 more answers
An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.
Ber [7]

Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

Explanation:

Given that,

Mass of first lead ball, m_1=5.2\ kg

Mass of the other lead ball, m_2=0.046\ kg

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N

So, the magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N. Hence, this is the required solution.

5 0
3 years ago
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