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Sedaia [141]
3 years ago
5

A physical science test book has a mass of 2.2 kg what is the weight on mars??? please help

Physics
1 answer:
Nastasia [14]3 years ago
4 0
Weight = (mass) x (gravity)
             = (2.2 kg) x (3.7 m/s^2)

             = 8.1 newtons. (about 1.84 pounds)
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The force between a pair of .005 charges is 750 N. What is the distance between them?
Ganezh [65]

Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?

Answer:

17.32 m

Explanation:

From coulomb's Law,

F = kqq'/r²........................... Equation 1

Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.

make r the subject of the equation above

r = √(kqq'/F)..................... Equation 2

From the question,

Given: q = q' = 0.005 C, F = 750 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(9.0×10⁹×0.005×0.005/750)

r = √(300)

r = 17.32 m.

Hence the distance between the pair of charges = 17.32 m

6 0
3 years ago
15. Missy Diwater, the former platform diver or the Ringling Brother's Circus had a speed of 25m/s when she hit the
timurjin [86]
K.E.= 1/2 x MV^2 = 1/2 x 40(kg) x (25x25) =12500J
5 0
2 years ago
This type of severe weather only develops when a group of storms meet over warm tropical waters what type of storm is this
erastovalidia [21]
Monsoon is the answer
5 0
3 years ago
What is the metric unit of weight? Gram/pound/Newton/milliliter
coldgirl [10]
Weight is force.
The metric unit of force is the Newton.
1 Newton is the force needed to accelerate 1 kilogram at the rate of 1 m/s² .
7 0
3 years ago
A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion j
nika2105 [10]

To solve this problem we will use the kinematic equations of descriptive motion of a projectile for which both the height reached and the distance traveled are defined. From this type of movement the lion reaches a height (H) of 3m and travels a horizontal distance (R) of 10 m. Mathematically the equations that describe this movement are given as,

H = \frac{v_0^2sin^2\theta}{2g}

R = \frac{v_0^2 sin 2\theta}{g}

Dividing the two equation we have that

\frac{H}{R}=\frac{\frac{v_0^2sin^2\theta}{2g}}{\frac{v_0^2 sin 2\theta}{g}}

\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{sin2\theta}

\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{2sin\theta cos\theta}

\frac{H}{R}= \frac{1}{4} \frac{sin\theta}{cos\theta}

\frac{H}{R}= \frac{1}{4} tan\theta

Substituting values of H and R, we get

\frac{3}{10} = \frac{1}{4} tan\theta

\theta = tan^{-1} \frac{12}{10}

\theta = 50.2\°

Substituting the value of \theta in equation we get,

H = \frac{v_0^2sin^2\theta}{2g}

v_0^2 = \frac{H 2g}{sin^2\theta}

v_0^2 = \frac{3*2*9.8}{sin^2(50.2)}

v_0^2 = 99.62

v_0 = \sqrt{99.62}

v_0 = 9.98m/s

Therefore the speed of the mountain lion just as it leaves the ground is 9.98m/s at an angle of 50.2°

5 0
3 years ago
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