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Dafna1 [17]
3 years ago
5

An object located near the surface of Earth has a weight of a 245 N

Physics
1 answer:
marusya05 [52]3 years ago
4 0

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

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A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction
steposvetlana [31]

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

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The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

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Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

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8 0
3 years ago
A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass
natka813 [3]

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet =  3.56×10²⁶ Kg.

5 0
3 years ago
In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give
sveta [45]

Answer:

The current pass the R_2 is  I  = 0.25 A

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  V =  3V

     The first resistance is  R_1 = 7 \Omega

     The second resistance is  R_2 = 5 \Omega

Since the resistors are connected in series their equivalent resistance is  

       R_{eq} =  R_1 +R_2

Substituting values

         R_{eq} = 7 + 5

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Since the resistance are connected in serie the current passing through the circuit  is the same current passing through R_2 which is mathematically evaluated as

        I  =  \frac{V}{R_{eq}}

Substituting values  

      I  =  \frac{3}{12}

      I  = 0.25 A

3 0
3 years ago
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