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baherus [9]
4 years ago
5

Dwight pulls his sister in her wagon with a force of 65 N at an angle of 50.0º to the ground. What are the magnitudes of the hor

izontal and vertical components of the force exerted by Dwight?

Physics
1 answer:
NeTakaya4 years ago
3 0
Refer to the diagram shown.

F = 65 N, the force exerted by Dwight.

The horizontal component of the force is
65 cos(50°) = 41.8 N
The vertical component of the force is
65 sin(50°) = 49.8 N

Answer:
Horizontal: 41.8 N
Vertical: 49.8 N


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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
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Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

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v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

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\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

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3 years ago
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Answer:

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Density = 1 x 10⁹ kg/m³

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4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)

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