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4 years ago

A motorcycle is traveling with a velocity of 4 m/s and has a mass of 112 kg

Physics

1 answer:

Zarrin [17]4 years ago

8 0

Answer: 896J

Explanation:

Given that

Mass of motorcycle = 112kg

kinetic energy = ?

velocity of the motorcycle = 4m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the motorcycle has kinetic energy

i.e K.E = 1/2mv^2

KE = 1/2 x 112kg x (4m/s)^2

KE = 0.5 x 112 x 16

KE = 896J

Thus, the motorcycle has 896 joules of kinetic energy.

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A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horiz

Rufina [12.5K]

Answer:

0.88752 kgm²

0.02236 Nm

Explanation:

m = Mass of ball = 1.2 kg

L = Length of rod = 0.86 m

$\theta$ = Angle = 90°

Rotational inertia is given by

$I=mL^2\\\Rightarrow I=1.2\times 0.86^2\\\Rightarrow I=0.88752\ kgm^2$

The rotational inertia is 0.88752 kgm²

Torque is given by

$\tau=FLsin\theta\\\Rightarrow \tau=2.6\times 10^{-2}\times 0.86sin90\\\Rightarrow \tau=0.02236\ Nm$

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3 years ago

Which of the following holds true for volume if the temperature of a given amount of gas increases and pressure remains constant

svlad2 [7]

D. Volume will increase
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7 0

3 years ago

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mylen [45]

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

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We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

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Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

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Mass of Saturn, Ms = 95.1 Me

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closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

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3 years ago

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Anuta_ua [19.1K]

Answer:im just guessing d but i think its d though

Explanation:

it pretty obvious

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3 years ago

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Nookie1986 [14]

Answer:

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Explanation:

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