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2 years ago

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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

upthrust on the body. (iii) volume of the body.

1 answer:

Lelechka [254]2 years ago

3 0

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

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Missy's favorite ride at the Topsfield Fair is the rotor, which has a radius of 4.0 m. The ride

dolphi86 [110]

Answer:

a. 12.57 m/s b. 39.5 m/s² c. Her centripetal force is four times her weight.

Explanation:

a. What is Missy's linear speed on the rotor?

Missy's linear speed v = 2πr/T where r = radius = 4.0 m and T = time it takes to complete one revolution = 2.0 s

So, v = 2πr/T

= 2π(4.0 m)/2.0 s

= 4π m/s

= 12.57 m/s

b. What is Missy's centripetal acceleration on the rotor?

Missy's centripetal acceleration, a = v²/r where v = linear velocity = 12.57 m/s and r = radius = 4.0 m

a = v²/r

= (12.57 m/s)²/4.0 m

= 158.01 m²/s² ÷ 4.0 m

= 39.5 m/s²

c. If her mass is 50-Kg, how is the centripetal force compare to her weight?

Her centripetal force F = ma where m = mass = 50 kg and a = centripetal acceleration = 39.5 m/s².

Her weight W = mg where m = mass = 50 kg and g = acceleration due to gravity = 9.8 m/s².

So, comparing her centripetal force to her weight, we have

F/W = ma/mg

= a/g

= 39.5 m/s² ÷ 9.8 m/s²

= 4.03

≅ 4

So her centripetal force is four times her weight.

8 0

3 years ago

30 points plus brainliest for best answer no funny answers Assignment: Waves Concept Map Exploration Your concept map will compa

mihalych1998 [28]

Answer:

Waves: "The phenomenon in which any number of particles or the energy packets that originates from a transmitter travels in a direction and it is somehow directed towards a receiver end is called as waves."

Electromagnetic wave: The type of waves which do not require any sort of medium for the transformation process of the energy, as it is directed towards from a transmitter towards a receiver.

For example: Radio, IR , Ultra sonic waves etc.

Mechanical waves: The type of waves which requires some medium to transfer the energy packets from one place to another, or from one point to the other using some medium. As the medium can be solids,liquids, or any gaseous forms.

For example: Sound waves, seismic waves etc.

Explanation:

The transfer of energy in Mechanical waves and Electromagnetic waves:

Both, the waves are much different due ones basic requirement of a medium while the other don't require one. As the density of any medium if increased it will increase the speed of the wave propagation. And this any medium being more denser will result in high speed form mechanical waves.

While, the electromagnetic waves can pass through matters or elements easily with out having any obstruction faced in its path because presence of medium is no requirement for the wave propagation.

I hope this helps!!

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3 years ago

When a 4.32 kg object is hung vertically on a certain light spring that obeys Hooke's Law, the spring stretches 2.92 cm.

Alika [10]

(a) 1.01 cm

First of all, we need to find the spring constant of the spring.

The force initially applied to the spring is equal to the weight of the block hanging on it:

$F=mg=(4.32)(9.8) = 42.3 N$

where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.

When this force is applied, the spring stretches by

$x=2.92 cm = 0.0292 m$

We can find the spring constant by using Hooke's law:

$F=kx$

where k is the spring constant. Solving for k,

$k=\frac{F}{x}=\frac{42.3}{0.0292}=1448.6 N/m$

Later, the first object is removed and another object of mass

m' = 1.50 kg

is hung on the spring. The weight of this object is

$F'=m'g=(1.50)(9.8)=14.7 N$

So, if we use Hooke's law again, we can find the new stretching of the spring:

$x'=\frac{F'}{k}=\frac{14.7}{1448.6}=0.0101 m = 1.01 cm$

(b) 1.16 J

The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:

$W=\frac{1}{2}kx^2$

where we have

k = 1448.6 N/m is the spring constant

x = 4.00 cm = 0.04 m is the new stretching

Solving the equation, we find the work that must be done by the external force:

$W=\frac{1}{2}(1448.6)(0.04)^2=1.16 J$

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2 years ago

What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9

mr_godi [17]

Answer:

3.2 m

Explanation:

The equation to use to solve this problem is:

$v_f^2 = v_i^2 + 2 a \Delta y$

where

$v_f$ is the final velocity

$v_i$ is the initial velocity

a is the acceleration

$\Delta y$ is the distance covered

For the particle in free-fall in this problem, we have

$v_i = 0$ (it starts from rest)

$v_f = 7.9 m/s$

$g=9.8 m/s^2$ (acceleration due to gravity)

By re-arranging the equation, we can find the distance travelled:

$\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m$

5 0

3 years ago

A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along

Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

$F(x)=(3\ N/m^2)x^2+(6\ N/m)x$

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

$W=\int\limits {F{\cdot} dx}$

$W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J$

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

3 0

3 years ago

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