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2 years ago

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The supporting force exerted by a fluid on an object immersed in it is called __________. Question 1 options: lift buoyant force

density viscosity

1 answer:

Dmitriy789 [7]2 years ago

8 0

The answer is buoyant force

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Magnetism is a ?<br> A) FORCE<br> B) POWER<br> C) FORM OF ENERGY<br> D) FORM OF ELECTRICITY

Ksivusya [100]

I believe this answer is Force

3 0

2 years ago

Read 2 more answers

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the

djverab [1.8K]

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.
It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

$a_{c} = \omega^{2} * r (1)$

Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.

∴ ωp = ωw (2)

⇒ $a_{p} = \omega_{p} ^{2} * r_{p} (3)$

$a_{w} = \omega_{w}^{2} * r_{w} (4)$

Dividing (4) by (3), from (2), we have:

$\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}$

Solving for aw, we get:

$a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2$

7 0

2 years ago

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

vladimir2022 [97]

Answer:

The distance is 300 m.

Explanation:

Given that,

Time = 30 s

Speed = 80 m/s

Distance = 1200 m

Speed of smaller plane = 40 m/s

We need to calculate the acceleration

Using equation of motion

$s= ut+\dfrac{1}{2}at62$

Put the value in the equation

$1200=0+\dfrac{1}{2}\times a\times(30)^2$

$a=\dfrac{2\times1200}{30\times30}$

$a=2.67\ m/s^2$

We need to calculate the distance

Using equation of motion

$v^2=u^2+2as$

Put the value in the equation

$40^2=0+2\times2.67\times s$

$s=\dfrac{40^2}{2\times2.67}$

$s=299.62\approx 300\ m$

Hence, The distance is 300 m.

3 0

2 years ago

A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters

emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

$F = \frac{mv^{2} }{r}$

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = $2\pi r$

Work on a particle = 2πrF = $2\pi mv^{2}$

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.

3 0

2 years ago

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × $10^{-26}$ N

(d) 7.37 × $10^{-29}$ N

Explanation:

(a) The minimum value of $x$ will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

$h^{2} = b^{2} + p^{2}$

$h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m$

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force= $F=\frac{kq1q2}{x^{2} }$

$F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }$

$F= 6.38$ × $10^{-26} N$

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F = $\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }$

F= 7.37× $10^{-29$ N

4 0

2 years ago