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Zinaida [17]
4 years ago
11

Gravity on Earth is 9.8 m/s2, and gravity on Mars is 3.7 m/s2.

Physics
2 answers:
antiseptic1488 [7]4 years ago
8 0
W = mg

Weight on Earth = 50 x 9.8
                          = 490 N

Weight on Mars = 50 x 3.7
                          = 185 N 
maria [59]4 years ago
6 0

Answer

Weight of the Jaden on earth is 490 N.

Weight of the Jaden on mars is 185 N.

Explanation:

Formula

w = mg

Where w is the weight of an object , m is the mass of an object and g is acceleration of gravity .

Case First

Gravity on Earth is 9.8 m/s² .

Jaden’s mass is 50 kilograms.

g = 9.8 m /s²

m = 50

Put the values in the formula

Thus

w = 50 × 9.8

   = 490 kg m /s²

(As N = kg m/ s²)

   = 490 N

Therefore the weight of the Jaden on earth is 490 N.

Case Second

Gravity on Mars is 3.7 m/s².

Jaden’s mass is 50 kilograms.

m = 50 kg

g = 3.7 m/s²

Putting the values in the formula

w = 50 × 3.7

   = 185 kg m/s²

(As N = kg m/ s²)

  = 185 N

Therefore the weight of the Jaden on mars is 185 N.


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A force

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Describe how a neutral material becomes attracted to a negatively charged object brought near it.
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It’s like static stuff

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Which combination of a wire coil and a core would make the weakest
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Read 2 more answers
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

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3 years ago
A ball rolls horizontally off a table of height 0.6 m with a speed of 9 m/s. How long does it take the ball to reach the ground?
denis23 [38]

Answer: 0.067 s

Explanation:s = Ut + 1/2at^2

0.6 = 9t + 0.5 *10 *t^2

Where a = g =10m/s/s

Solving the quadratic equation

5t^2 + 9t - 0.6=0,

t= 0.067 s and - 1.7 s

Of which 0.067 s is a valid time

4 0
3 years ago
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