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Hunter-Best [27]
3 years ago
12

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (

Note: 1 kg = 1000 g).
Physics
1 answer:
slavikrds [6]3 years ago
6 0

Answer:

Momentum, p = 7.2 g-m/s

Explanation:

It is given that,

The momentum of an object is p=7.2\times 10^{-3}\ kg-m/s

We need to express momentum in any equivalent units. There can be many solutions of this problem. Some of the units of mass are gram, milligram etc. units of length are meters, mm etc.

Since, 1 kg = 1000 gram

So, p=7.2\times 10^{-3}\times 10^3\ g-m/s

Therefore, the momentum of the object is 7.2 g-m/s. Hence, this is the required solution.

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Resistance = (voltage) / (Current)

Resistance = (10 V) / (5 A)

Resistance = 2 ohms.

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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
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}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

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.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

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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A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The
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Answer:

the magnitude of acceleration will be 1.50m/s^2

Explanation:

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if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

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you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

Fnet=ma

(15.0N)=(10.0kg)a    from her you can divide both sides by 10 to isolate a:

1.50=a  (and now make sure to label the units of your answer)

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Both electric and magnetic fields
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