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Elena-2011 [213]
4 years ago
8

Acceleration of gravity on jupiter

Physics
2 answers:
Viefleur [7K]4 years ago
5 0

The acceleration of gravity on Jupiter is listed as  <em>24.79 m/s²</em> .

That's roughly  2.53  times its value on Earth.  So if you weigh, let's say,
130 pounds on Earth, then you would weigh about 328 pounds on Jupiter.



AlladinOne [14]4 years ago
4 0
Jupiter acceleration of gravity is approximately 26 (m/s) / s.
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In which of the following cases will the normal force on a box be the greatest? A. When the box is placed in a stationary elevat
Butoxors [25]

Answer:

 D. When the box is placed in an elevator accelerating upward

Explanation:

Looking at the answer choices, we know that we want to find out how the normal force varies with the motion of the box. In all cases listed in the answer choices, there are two forces acting on the box: the normal force and the force of gravity. These two act in opposite directions: the normal force, N, in the upward direction and gravity, mg, in the downward direction. Taking the upward direction to be positive, we can express the net force on the box as N - mg.

From Newton's Second Law, this is also equal to ma, where a is the acceleration of the box (again with the upward direction being positive). For answer choices (A) and (B), the net acceleration of the box is zero, so N = mg. We can see how the acceleration of the elevator (and, hence, of the box) affects the normal force. The larger the acceleration (in the positive, i.e., upward, direction), the larger the normal force is to preserve the equality: N - mg = ma, N = ma+ mg. Answer choice (D), in which the elevator is accelerating upward, results in the greatest normal force, since in that case the magnitude of the normal force is greater than gravity by the amount ma.

4 0
3 years ago
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A ball is throw into the sky at 18m/s. How high can it reach?
Komok [63]

Answer:

The maximum height of the ball will be "16.53 m".

Explanation:

Given that,

V_{final} = 18 \ m/s

As we know,

The initial as well the final height on the ball will be same after it reaches it's maximum height.

then,

⇒ V_{final}=V_{initial}

where, V_{initial} = \sqrt{2gh}

                       =\sqrt{2\times 9.8\times h}

                       =\sqrt{19.6h}

hence,

The max height of the ball will be:

⇒ 18=\sqrt{19.6h}

    19.6h=324

           h=\frac{324}{19.6}

              =16.53 \ m

8 0
3 years ago
Suppose the experiment was conducted in the same manner, but the axle was now off center of the solid disk. Would you expect the
san4es73 [151]

Answer:

Explanation:

If Ig be moment of inertia about an axis through centre of mass and I be moment of inertia through any other axis parallel to earlier axis , then according to theory of parallel axis ,

I = Ig + Md²

where M is mass of the body and d is distance between two parallel axis.

So I is greater than Ig.

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3 years ago
What is the only continent with land in all four hemispheres?
dimaraw [331]
It’s either Asia or the America
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3 years ago
Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
4 years ago
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