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Radda [10]
3 years ago
7

The mass of a proton is approximately equal to

Physics
1 answer:
Dahasolnce [82]3 years ago
3 0

Answer:

1.6726\cdot 10^{-27} kg

Explanation:

The three main particles that make an atom are:

- Proton: its mass is 1.6726\cdot 10^{-27} kg, it carries an electric charge of +e (e=1.6\cdot 10^{-19}C), and it is located in the nucles of the atom

- Neutron: its mass is 1.6749 \cdot 10^{-27}kg, it carries no electric charge, and it is also located in the nucleus of the atom

- Electron: its mass is 9.1094 \cdot 10^{-31}kg, it carries an electric charge of -e (e=1.6\cdot 10^{-19}C), and it is located outside the nucleus

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Ohms law is A.(R=E/W). B.(R=E/1). C.(E/Z). D.none of them
nikklg [1K]

Answer:

D. none of them.

Explanation:

This is because Ohm's law is:

Voltage = Current × Resistance

or,

V = IR

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2 years ago
Convert 400 mm to m using the method of dimensional analysis
s2008m [1.1K]

Answer:

To convert 400 mm to m you can apply the formula [m] = [mm] / 1000; use 400 for mm. Thus, the conversion 400 mm m is the result of dividing 400 by 1000. 0.4

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2 years ago
If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?
Alex

Answer:

0.015 m/s2

Explanation:

Using Newtons 2nd law.

F = ma where F = Force applied, m = mass of the object and a = acceleration acquired.

So substitute the values in SI units.

m = \frac{3}{1000} kg

Therefore F = 0.003×5 = 0.015 m/s2

3 0
2 years ago
which of the following are vector quantities? check all that apply. a. force b. acceleration c. displacement d. mass
Ratling [72]
Force, acceleration, and Displacement are all vector quantities.
4 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
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