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Tasya [4]
3 years ago
8

Gravity on Jupiter is 25 m/s/s, what is the weight of a 12 kg object on Jupiter?

Physics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

300 Newtons

Explanation:

Weight is the force of attraction between two bodies, one usually larger (like a planet), and one smaller (like a person). Force can be calculated using the formula: Force = mass × acceleration.

The mass here is 12kg, the acceleration, which in this case, is the acceleration due to gravity is 25m/s/s, by plugging in our values, we have

Force = 12 × 25 = 300 Newtons or 300 N for short.

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Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us as
Natali [406]

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

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3 years ago
How is the average kinetic energy of particles related to temperature of a substance?
zubka84 [21]
As kinetic energy increases, substance temperature increases
6 0
3 years ago
Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was
tigry1 [53]

Answer:

\Delta V=316167V

Explanation:

The difference of electric potential between two points is given by the formula \Delta V=Ed, where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is E=\frac{F}{q}, where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have \Delta V=\frac{Fd}{q}, so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force N=W-F. On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) d=\frac{at^2}{2}, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

a_1=\frac{2d}{t_1^2}

a_2=\frac{2d}{t_2^2}

These relate with the forces by Newton's 2nd Law:

W=ma_1

N=ma_2

Putting all together:

N=W-F=ma_1-F=ma_2

Which means:

F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})

And finally we substitute:

\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})

Which for our values means:

\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V

7 0
3 years ago
A block weighing 3.7 kg is suspended from the ceiling of a truck trailer by a hanging bungee cord. The cord has a cross-sectiona
sweet [91]

Answer:

Y = 2.27 \times 10^{10} N/m^2

Explanation:

Natural length of the string is given as

L_o = 43 cm

length of the string while block is hanging on it

L = 53 cm

extension in length is given as

\Delta L = 10 cm

now we have strain in the string is given as

strain = \frac{\Delta L}{L}

strain = \frac{{10 cm}{43 cm}

strain = 0.23

similarly we will have cross-sectional area of the string is given as

A = 40 \times 10^{-6} m^2

now the stress in the string is given as

Stress = \frac{T}{A}

Stress = \frac{mg}{A}

Stress = \frac{3.7 \times 9.81}{40 \times 10^{-6}}

stress = 9.07 \times 10^5 N/m^2

Now Young's Modulus is given as

Y = \frac{stress}{strain}

Y = \frac{9.07 \times 10^5}{40\times 10^{-6}}

Y = 2.27 \times 10^{10} N/m^2

5 0
3 years ago
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