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3 years ago

What do we call the center of an atom that contains the mass of the with ca

Physics

1 answer:

Delvig [45]3 years ago

6 0

Answer:

Hope this helps! Please Mark Brainliest!

Explanation:

The Nucleus: The Center of an Atom. The nucleus, that dense central core of the atom, contains both protons and neutrons. Electrons are outside the nucleus in energy levels.

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what is the force if a block of wood with mass 20 kg slides along a frictionless surface at 2 m/s squared

CaHeK987 [17]

Answer:

40 N

Explanation:

F=ma where F is the applied force, m is the mass of object and a is the acceleration.

Since there is no friction, substituting 20 Kg for m and 2 m/s squared for a then we obtain

F=20*2=40 N

5 0

3 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

7 0

3 years ago

What is Kepler’s first law?

BartSMP [9]

Answer:

Kepler's First Law: each planet's orbit about the Sun is an ellipse. The Sun's center is always located at one focus of the orbital ellipse. The Sun is at one focus. The planet follows the ellipse in its orbit, meaning that the planet to Sun distance is constantly changing as the planet goes around its orbit.

5 0

2 years ago

Read 2 more answers

A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of

Roman55 [17]

Answer:

$T=9.42Nm$

Explanation:

From the question we are told that:

Radius $r= 0.5 m$

Current $I= 3.0 A$

Normal vector $n=\frac{(2i - j +2k)}{3}$

Magnetic field $B= (2i-6j) T$

Generally the equation for Area is mathematically given by

$A=\pi r^2$

$A=3.1415 *0.5^2$

$A=0.7853 m^2$

Generally the equation for Torque is mathematically given by

$T=A(i'*B)$

Where

$i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}$

$X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]$

$X\ component\ of\ i'*B=12$

Therefore

$T=0.7853*12$

$T=9.42Nm$

7 0

3 years ago

1 example of Deionization

xxTIMURxx [149]

Answer:

For example, the rain water does not have any salinity, except the small quantity due to the atmosphere, while the sea water has a very high salinity.

hope this helps you friend

7 0

3 years ago