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3 years ago

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When a particle of mass m is at (x comma 0 ), it is attracted toward the origin with a force whose magnitude is StartFraction y

Over x squared EndFraction . If the particle starts from rest at x equals b and is acted upon by no other forces, find the work done on it by the time it reaches x equals a, 0 less than a less than b.

1 answer:

Dahasolnce [82]3 years ago

8 0

Answer:

W = y (b-a) / ab

Explanation:

Work is defined by the expression

W = ∫ F. dr

In this case the force is in the same direction of displacement, so the scalar product is reduced to the ordinary product

W = ∫ F dr

The expression of the strength left is

F = -y / x²

let's replace and integrate

W = ∫ (-y / x²) dx

W = -y (-1 / x)

We evaluate between the lower limit x = b + a to the upper limit x = 0 + a

W = -y (-1 / b + 1 / a)

W = y (b-a) / ab

where (b-a) is the distance traveled

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Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet

telo118 [61]

Answer:

- Bigger mass of planet B

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

$v=\sqrt{\frac{GM}{r}}$

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B

- orbiting closer to planet B

6 0

3 years ago

A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the

Lelechka [254]

Answer:

$difference \: in \: weight = 150n - 100n = 50n$

Now,buyantant force

$difference \: in \: weight \: = volume(body) \times density \: of \: water \: \times g$

so;

$50 = {v}^{b} \times 1 \times {10}^{3} \times 9.8m {s}^{2}$

${v}^{b} = \frac{50}{1000 } \times 9.8$

$= \frac{50}{9800}$

$= 0.0051$

Now,

$mass \: in \: air \: = 150n = \frac{150}{9.8kg}$

$density = \frac{weght}{volume}$

$= \frac{150}{0.0051} \times 9.8 \\ x = 3000$

And now,

$specific \: density \: = \frac{density of \: the \: body}{density \: of \: water}$

$= \frac{3000}{1000}$

$= 3$

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0

3 years ago

Por una resistencia de 10 Ω fluyen 5A. ¿Cuál será la diferencia de potencial que se le debe aplicar a la resistencia?

ra1l [238]

Answer:

V = 50 volts

Explanation:

Given that,

Resistance, R = 10 ohms

Current, I = 5 A

We need to find the potential difference across the circuit. We know that,

V = IR

Put all the values,

V = 5 × 10

V = 50 volts

Hence, the potential difference is equal to 50 volts.

8 0

3 years ago

If the work put into a lever is 930 joules and the work accomplished is 870 joules, the efficiency of the lever is _____.

77julia77 [94]

♥ C) 94%
♥ If the work put into a lever is 930 joules and the work accomplished is 870 joules, the efficiency of the lever is 94%.
♥ <span>870/930=93.5
</span>♥ And rounded you get 94.

5 0

3 years ago

Read 2 more answers

A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th

Stolb23 [73]

Answer:

The current is $I_b = 400 \ A$

Explanation:

From the question we are told that

The length of the segment is $l = 2.50 \ m$

The current is $I_a = 1000 \ A$

The force felt is $F = 4.0 \ N$

The distance of the second wire is $d = 5.0 \ cm = 0.05 \ m$

Generally the current on the second wire is mathematically represented as

$I_b = \frac{2 \pi * r * F }{ l * \mu_o * I_a }$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4 \pi * 10^{-7} \ N/A^2$

=> $I_b = \frac{2 * 3.142 * 0.05 * 4 }{ 2.50 * 4\pi *10^{-7} * 1000 }$

=> $I_b = 400 \ A$

4 0

3 years ago