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Furkat [3]
3 years ago
15

Determine the position in which a solid cylindrical block of wood of diameter 0.3 m and length 0.4 m will float in water. Take s

pecific gravity of wood as 0.5
Physics
1 answer:
yan [13]3 years ago
4 0

Answer:

0.2 m

Explanation:

Diameter = 0.3 m

radius, r = 0.15 m

Length, H = 0.4 m

density of wood, d = 0.5 g/cm^3 = 500 kg/m^3

density of water, d = 1000 kg/m^3

Let h be the depth of cylinder immersed in water.

By the principle of floatation.

Buoyant force = Weight of cylinder

Volume immeresed x density of water x g = Volume of cylinder x density of wood x g

A x h x 1000 x g = A x H x 500 x g

1000 h = 500 x 0.4

h = 0.2 m

You might be interested in
Machmer Hall is 400 m North and 180 m West of Witless.
yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

D = \sqrt{(d_{m})^2+(d_{w})^2}

Where, d_{m} =distance of Machmer Hall

d_{w} =distance of Witless

Put the value into the formula

D = \sqrt{(400)^2+(180)^2}

D=438.63\ m

Hence, The distance from Witless to Machmer is 438.63 m.

5 0
3 years ago
In order for an object to have kinetic energy it must have a mass and a ?
almond37 [142]

Answer:

Velocity

Explanation:

  • The mechanical energy of the body is defined as the sum of the potential energy and kinetic energy.

                                   E = P.E + K.E

  • The potential energy of a body is due to the height from the surface of the earth.

                                  P.E = mgh

  • The kinetic energy of the is possessed by the body due to the virtue of its motion,

                                  K.E = ½ mv²

  • If there is no velocity associated with the body, there is no K.E in the body.
8 0
3 years ago
An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)
Alecsey [184]
"60 kg" is not a weight.  It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is   

                                 (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

                   Weight = (60 kg) x (9.8 m/s²)

                                =      588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to  5R  from the center, the gravity out there is

                    (1R/5R)²  =  (1/5)²  =  1/25  =  0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

                 (0.04) x (588 Newtons)  =  23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink. 
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
6 0
3 years ago
The density of silver is 10.5 g cm3. a piece of silver with a mass of 61.3 g would ovvupy a volume of ?
natali 33 [55]
The relationship between mass m, volume V and density d is:
d= \frac{m}{V}
The silver has density d=10.5 g/cm^3, and the mass of the piece of silver is m=61.3 g. Therefore we can calculate its volume using the previous formula:
V= \frac{m}{d}= \frac{61.3 g}{10.5 g/cm^3}=5.84 cm^3
7 0
3 years ago
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th
Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:  

V_{f}^{2} = V_{0}^{2} - 2gh

<u>Where:</u>

V_{f}: is the final speed = 14 m/s

V_{0}: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s  

b) The maximum height is:

V_{f}^{2} = V_{0}^{2} - 2gh

h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m

c) The time can be found using the following equation:

V_{f} = V_{0} - gt

t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s

d) The flight time is given by:

t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s

         

I hope it helps you!    

3 0
3 years ago
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