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VLD [36.1K]
4 years ago
11

A 2.00-kilogram object weighs 19.6 newtons on earth. if the acceleration due to gravity on mars is 3.71 meters per second2, what

is the object’s mass on mars
Physics
1 answer:
Harrizon [31]4 years ago
4 0
Recall that mass is the amount of matter present in a body. That means it's a property that is consistent regardless of the body's current location, gravity's pull on the body, etc. 

Let's not confuse mass with weight (which is a force computed as Weight = mass x acceleration). Mass will remain constant and that means that whether the object is on Earth or on Mars, its mass remains the same.  Thus, the object will still have 2.00 kg as mass on Mars. 

Answers: 2.00 kilograms
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Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S an
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Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

F=k_e\frac{q_1q_2}{r^2}

where,

F = electrostatic force of attraction or repulsion

k_e = Coulomb's constant

q_1 and q_2 are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

F_{sq}=k_e\frac{qq}{1^2}=k_e\times q^2       ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

F_{sp}=k_e\frac{qq}{2^2}=k_e\frac{q^2}{4}     ...........(2)

Equation equation 1 and 2, we get

\frac{F_{sq}}{F_{sp}}=\frac{1}{4}

F_{sq}=4\times F_{sp}

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

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3 years ago
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Answer:

The time it takes to stop is 13.75 seconds

Explanation:

A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;

v = u + a·t

Where;

v = The final velocity of the body

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u = The initial velocity of the body

The given parameters for the acceleration of the car are;

The initial velocity of the car, u = 0 m/s (a car starting from rest)

The constant acceleration of the car, a =  5.5 m/s²

The acceleration duration, t = 6 s

Therefore, we have;

The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s

The final velocity of the car after the acceleration, v = 33 m/s

When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'

The given parameters when the car slows down  are;

The deceleration, -a₂ = 2.4 m/s²

The final velocity, v₂ = 0 m/s

The initial velocity, u₂ = v = 33 m/s

The time it takes to stop = t₂

-a₂ = 2.4 m/s²

∴ a₂ = -2.4 m/s²

From, v = u + a·t, we have;

v₂ = v + a₂·t₂

By plugging in the values of the variables, we have;

0 m/s = 33 m/s + (-2.4 m/s²) × t₂

∴ 2.4 m/s² × t₂ = 33 m/s

t₂ = 33 m/s/(2.4 m/s²) = 13.75 s

The time it takes to stop, t₂ = 13.75 seconds

7 0
3 years ago
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