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3 years ago

9

An student in a swimming pool weighs 450 N and in a regular balance out of the swimming pool weighs 700 N. What is the value of

the buoyant force?

1 answer:

rewona [7]3 years ago

7 0

The value of buoyant force is 700-450=300N

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What do we call the distance labeled from A to B and what could we do to the note played on an instrument to change that distanc

Korvikt [17]

Correct answer choice is :

A) From A to B is known as the wavelength and changing the pitch of the note will change its length.

Explanation:

The amount or quantity of period within two things, points, lines, etc. the state or fact of existing separate in space, as of one thing from another; remoteness. a linear amount of space: Seven miles is a distance too great to walk in an hour. Distance is a scalar quantity describing the interval in two points. It is just the measure of the interval.

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Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

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

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

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

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of <u>2.1 m</u>

7 0

3 years ago

The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx = -6.1 and By = -2.2. W

Zina [86]

For simplicity, let's call vector B-A vector C Then C is
Cx = (-6.1 - 2.2)
Cy = (-2.2 - (-6.9)) Or,
Cx = -8.3 Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538

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3 years ago