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3 years ago

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g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force a

cting on the stone throughout its flight is constant, independent of the velocity of the stone, and has a magnitude of 0.900 N. what is the maximum height reached by the stone?

1 answer:

otez555 [7]3 years ago

5 0

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

$-mgh - F_f h = 0 - \frac{1}{2}mv_i^2$

now we will have

$-1.60(9.8)(h) - 0.900(h) = - 470$

$-16.58 h = -470$

$h = 28.35 m$

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction

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</span>Yes. As we said in point b, the magnetic force is
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3 years ago

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Answer:

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Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

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${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

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${v_{f}=\sqrt{2*9.81*2.41$

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Given Information:

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Answer:

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Explanation:

The mass of the planet can be found using Newton's law of gravitation,

g = GM/R²

M = gR²/G

Where g is the gravitational acceleration at the planet, R is the radius of the planet and G is the gravitational constant.

G = 6.6743×10⁻¹¹ m³/kg⋅s²

First we need to find R and g

The relation between circumference and radius is given by

C = 2πR

Where R is the radius of planet

25.7 = 2πR

R = 25.7/2π

R = 4.09 km

From the equations of motion we have,

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Where u is the initial speed which is zero, t is the time and h is the height of release.

Re-arrange the above equation for g

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g = 0.003177 m/s²

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M = 0.003177*(4.09×10³)²/6.6743×10⁻¹¹

M = 7.963×10¹⁴ kg

Therefore, the mass of the given planet is 7.963×10¹⁴ kg

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