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taurus [48]
3 years ago
15

An object of mass 100g projected vertically upwards from the ground level has a velocity of 20m/s at a height of 10m. Calculate

the initial kinetic energy at the ground level.​
Physics
1 answer:
posledela3 years ago
4 0

Answer:

kinetic energy = 20 J

Explanation:

KE = ½mv²

= ½× <u>100 </u> × 20 × 20

1000

KE = 20 J//

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(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and dia
stealth61 [152]

Answer:

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = 6.37\times 10^{6}\ m

g = Acceleration due to gravity = 9.81 m/s²

\omega = Angular speed

Force at equator would be

F_e=m(g-\omega^2R)

Force at pole

F_p=mg

From the question

F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}

Time period is given by

T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours

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5 0
3 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

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\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

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