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3 years ago

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

Physics

2 answers:

antiseptic1488 [7]3 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]3 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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Answer:

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2 years ago

Calculate the amount of charge necessary to exert a force of 35 N in an electric field of 300 N/C.

densk [106]

Answer:

The amount of charge is "0.117 C" i.e., option A.

Explanation:

The given values are:

Force,

$\vec{F}$ = 35 N

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Charge,

q = ?

As we know,

⇒ $\vec{F}=q\vec{E}$

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3 years ago

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3 years ago

If the work required to speed up a car from 11 km/h to 21 km/h is 6.0×103 J , what would be the work required to increase the ca

goblinko [34]

Explanation:

We need convert the velocities first to m/s and we get the following:

v2 = 21 km/hr = 5.8 m/s

v1 = 11 km/hr = 3.1 m/s

We need to find the mass of the car also for later use do using the work-energy theorem:

$delta \: w = \frac{1}{2} m(v \frac{2}{2} - v\frac{2}{1} )$

6.0x10^3 J = (0.5) m [(5.8)^2 - (3.1)^2]

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Now we determine work needed delta W to change its velocity from 21 km/hr to 33 km/hr

v2 = 33 km/hr = 9.2 m/s

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6 0

3 years ago

The moon's mass is 7.35 × 1022 kg, and it moves around the earth approximately in a circle or radius 3.82 × 105 km. The time req

Stolb23 [73]

Explanation:

It is given that,

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The time required for one revolution is 27.3 days, t = 27.3 days

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27.3 days = 2358720 seconds

Let v is the speed of moon around the circular path. It is given by :

$v=\dfrac{2\pi r}{T}$

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v = 1017.57 m/s

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$F=\dfrac{mv^2}{r}$

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3 years ago