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Zinaida [17]
3 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]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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Con que nombre se conocieron los primeros gimnasio que se crearon​
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Answer:

Los primeros gimnasios registrados datan de hace más de 3000 años en la antigua Persia, donde se los conocía como zurkhaneh, áreas que fomentaban la aptitud física.

7 0
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

Electric field,

\vec{E} = 300 N/C

Charge,

q = ?

As we know,

⇒  \vec{F}=q\vec{E}

On substituting the given values in the above formula, we get

⇒  35=q\times 300

⇒   q=\frac{35}{300}

⇒   q=0.117 \ C

3 0
3 years ago
If you add 700 kJ of heat to 700 g of water at 70 degrees C, how much water is left in the container? The latent heat of vaporiz
makkiz [27]

Answer:A

Explanation:Find attached picture file for details

3 0
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]

or

m = 499.4 kg

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

v1 = 21 km/hr = 5.8 m/s

delta W = (0.5)(499.4)[(9.2)^2 - (5.8)^2]

= 1.3 x 10^4 J

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,

Mass of moon, m=7.35\times 10^{22}\ kg

Radius of circle, r=3.82\times 10^{5}\ km=3.82\times 10^{8}\ m

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

1 day = 86400 seconds

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}

v=\dfrac{2\pi \times 3.82\times 10^{8}\ m}{2358720\ s}

v = 1017.57 m/s

Let F is the centripetal force acting on the moon. It is given by :

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

F=\dfrac{7.35\times 10^{22}\ kg\times (1017.57\ m)^2}{3.82\times 10^{8}\ m}

F=1.99\times 10^{20}\ m/s^2

So, the centripetal force that must act on the moon is 1.99\times 10^{20}\ m/s^2. The gravitational force that the earth exerts on the moon at that same distance is also equal to 1.99\times 10^{20}\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
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