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lesya [120]
3 years ago
13

A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of al

uminum is 2698.4 kg / m3
Physics
1 answer:
Illusion [34]3 years ago
7 0

Answer:

W =1562.53 N

Explanation:

It is given that,

Radius of the aluminium ball, r = 24 cm = 0.24 m

The density of Aluminium, d=2698.4\ kg/m^3

We need to find the thrust and the force. The mass of the liquid displaced is given by :

m=dV

V is volume

Weight of the displaced liquid

W = mg

W=dVg

So,

W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N

So, the thrust and the force is 1562.53 N.

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A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What
tangare [24]

Answer:

man will move in opposite direction with speed

v_1 = 1.66 \times 10^{-3} m/s

Explanation:

As we know that man is lying on the friction-less surface

so here net force along the surface is zero

so if we take man + stone as a system then net change in momentum of this system will become zero

so here we have

P_i = P_f

0 = m_1v_1 + m_2v_2

here we have

0 = (97)v_1 + 0.062(2.6)

v_1 = -\frac{0.1612}{97}

v_1 = -1.66 \times 10^{-3} m/s

3 0
3 years ago
What can be found in every skeletal muscle?
anastassius [24]
You can find muscle fibers, nerves, connective tissue, and blood vessels in every skeletal system. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.
7 0
3 years ago
Read 2 more answers
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s
IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

r = \frac{a(1-E^2)}{1+Ecos\beta }   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

\beta = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

v^2 = \frac{4\pi^2 }{r_{c} }   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

(V^2 = (\frac{4\pi^{2}  }{149.626*10^9})

therefore : V = 1.624* 10^-5 m/s

4 0
4 years ago
Which statement best describes how electric fields and magnetic fields are alike?
gladu [14]
<span>Electric field repulsive for objects of like charge and attractive for opposite type of charges and for a magnet you can say that like poles repel and unlike attracts so D makes sense</span>
3 0
3 years ago
A student was traveling to see his grandmother who lives 15 miles north oh his home he started from rest and maintained a pace o
N76 [4]

Explanation:

Given parameters:

Distance  = 15miles north = 24140.2m

Initial velocity  = 0m/s

Final velocity  = 4m/s

Unknown:

Speed, velocity and acceleration = ?

Solution:

The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.

  Speed  = \frac{distance }{time}  

  The speed of the student is 4m/s

Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;

    Velocity  = \frac{displacement}{time}  

   The velocity of the student is 4m/s due north

Acceleration is the change in velocity with time;

     To find the acceleration, we use

      v²  = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

     4² = 0² + 2x a x 24140.2

       a  = \frac{16}{2 x 24140.2}  = 0.00033m/s²

3 0
3 years ago
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