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aniked [119]
3 years ago
8

The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will

have what mass and weight on Jupiter?
Physics
1 answer:
inessss [21]3 years ago
5 0

Answer:  weight on Jupiter = 869.75 N

              mass  on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁  = 347,9 N

On the Jupiter,

G₂ = mg₂  

mass on the Earth = mass on the Jupiter !  

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N =  869,75 N

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There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal
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Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2  x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75  x 42 / 3

= 1050 Ns.

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3 years ago
A cardio kickboxing class challenges 3 things. List them.
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Answer:

In a cardio kickboxing class, you will learn proper form for the famous jab, hooks,crosses, uppercuts, back kicks, front kicks.

Explanation:

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Am 1 years old UwU (JK 15)
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A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu
seropon [69]

Answer:

\theta=50\ revolution

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, \omega_i=600\ rpm=62.83\ rad/s

Finally it shuts off, \omega_f=0

Time taken, t = 10 s

Let \alpha is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{0-62.83}{10}

\alpha =-6.283\ rad/s^2

Let \theta is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }

\theta=\dfrac{-62.83^2}{2\times -6.283}

\theta=314.15\ radian

\theta=49.99\ revolution

or

\theta=50\ revolution

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0
3 years ago
A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor
Kaylis [27]
<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span>U = R*i → i =  \frac{U}{R}<span>

Solving:

</span>i = \frac{U}{R}
i = \frac{12}{5}
\boxed{i = 2.4A}

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>
5 0
3 years ago
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