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2 years ago

Why are scientific models important ?

Physics

2 answers:

Sindrei [870]2 years ago

7 0

They help scientists explain concepts that are difficult to observe.

ZanzabumX [31]2 years ago

6 0

It depends on the situation but to many they are important because they help visualize things that are very complex.

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Two point charges are placed on the x-axis as follows: charge q1 = 3.95 nC is located at x= 0.198 m , and charge q2 = 4.96 nC is

Yuki888 [10]

Answer:

F = 2.40 × $10^{-6}$ N

Explanation:

given data

charge q1 = 3.95 nC

x= 0.198 m

charge q2 = 4.96 nC

x= -0.297 m

solution

force on a point charge kept in electric field F = E × q ................1

here E is the magnitude of electric field and q is the magnitude of charge

and

first we will get here electric field at origin

So net field at origin is

E = (Kq2÷r2²) - (kq1÷r1²) ...............2

put here value

E = 9[(4.96÷0.297²)-(3.95÷0.198²)]

E = 400.72 N/C ( negative x direction )

so that force will be

F = 6 × $10^{-9}$ × 400.72

F = 2.40 × $10^{-6}$ N

7 0

3 years ago

TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the po

Ghella [55]

Answer:

The speed is $138852.4 \frac{m}{s}$

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

$F=Eq$

with q the charge of the proton ( $1.61\times10^{-19} C$ ). Because between parallel plates electric field is almost constant, electric force is constant too:

$F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N$

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

$a=\frac{F}{m}$

with m the mass of the proton ( $1.67\times10^{-27} kg$ ):

$a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}$

Now with this constant acceleration we can use the kinematic equation

$v^{2}=v_{0}^{2}+2ad$

with v the final speed, $v_i$ the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

$v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)}$

$v=138852.4 \frac{m}{s}$

5 0

3 years ago

The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t

Vika [28.1K]

Answer:

The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of average pressure

$P_{avg}=\rho\times g\times d_{avg}$

Put the value into the formula

$P_{avg}=1000\times9.8\times100$

$P_{avg}=980000\ Pa$

We need to calculate the hydro static force on the back of the dam

Using formula of force

$F = P_{avg}\times A$

Put the value into the formula

$F = 980000\times1000\times200$

$F=1.96\times10^{11}\ N$

Hence, The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

7 0

2 years ago

Without the wheels, a bicycle frame has a mass of 8.29 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit

trapecia [35]

Answer:

69.66 Joule

Explanation:

mass of bicycle frame, mf = 8.29 kg

mass of wheel, mw = 0.820 kg

radius, r = 0.343 m

velocity, v = 3.6 m/s

There are two wheels in the bicycle.

There are two types of kinetic energy of the system one is kinetic energy of rotation and another is rotational kinetic energy.

$K = \frac{1}{2}m_{f}v^{2}+ 2\times \frac{1}{2}m_{w}v^{2}+ 2\times \frac{1}{2}I_{w}\omega^{2}$

$K = \frac{1}{2}m_{f}v^{2}+m_{w}v^{2}+ \frac{1}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}m_{f}v^{2}+ \frac{3}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}\times 8.29\times 3.6^{2}+ \frac{3}{2}\times 0.820\times 3.6^{2}$

K = 69.66 J

3 0

3 years ago

A Lamborghini Aventador engine reaches its top speed from rest in 3 seconds. The engine performs 1,545,000 Joules of work in tha

nikitadnepr [17]

Answer:

W = 1.545E6 J total work

P = W / t = 1.545E6 J / 3 sec = 5.15E5 J/sec = 515,000 J/sec (Watts)

Using definition of power

5 0

2 years ago