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krok68 [10]
3 years ago
15

A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on Mars?

Physics
2 answers:
zhannawk [14.2K]3 years ago
8 0
Field strength =force/unit mass
= 14.8N/ 4kg = 3.7N/kg
Shalnov [3]3 years ago
8 0

Answer:

The gravitational field strength on Mars is 3.7 N/kg.

Explanation:

Given that,

Mass of the object, m = 4 kg

Weight of the object on Mars, F = W = 14.8 N

We need to find the gravitational field strength on Mars. It is given by gravitational force per unit mass of an object. Mathematically, it is given by

G=\dfrac{F}{m}

G=\dfrac{14.8\ N}{4\ kg}

G = 3.7 N/kg

So, the gravitational field strength on Mars is 3.7 N/kg. Hence, this is the required solution.

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madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy =\frac{1}{2} mv^2, where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0
3 years ago
I need help with my physics homework agh! Please help it's due tomorrow. <br>​
storchak [24]

Rubbing both pieces cause each piece to have a negative charge.

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If they held the piece to the other end of the one held by a string it would start to rotate in the opposite direction.

4 0
3 years ago
Compare the force of air resistance and the force of gravity on an object falling at its terminal velocity.
Sholpan [36]
The viscous force on an object moving through air is proportional to its velocity.
The only forces acting on an object when falling are air resistance and its weight itself. The weight acts vertically downwards whereas air resistance acts vertically upward.
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7 0
4 years ago
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
3 years ago
What is meant by the statement,the linear expansivity of copper is 0.000017k
Aliun [14]

Answer:

The change in length per unit length per degree rise in temperature of copper is 0.000017k

Explanation:

Given that :

The linear expansivity of copper is 0.000017k. This simply means that ; for a given copper length, the length of such copper will increase by 0.000017k for every degree rose in temperature of the copper rod.

Therefore, the change in length per unit length per degree rise in temperature (k) is 0.000017

4 0
3 years ago
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